Compute the value of the sum $7cdot 11+11cdot 15+15cdot 19+cdots+95cdot 99$












0












$begingroup$


Solution: My attempt:
=$9^2-2^2 +13^2-2^2+17^2-2^2...97^2-2^2$

=$(9^2+13^2+17^2...97^2)-2^2( 23 )$
Focus on the first part. 81+169+289+441...Here first-order differences are 88, 120, 152... and the second-order difference is common to be 32. Therefore the series is of the form of $p(n)^2+q(n)+r$. If we plug n=0, we get the first term, if plug n=1, we get the second term, if we plug n=2, we get the third term and so on.
$p*0^2+q*0+r=81$
r=81
$p*1^2*q*1+81=169$
$p+q+81=169$
$p+q=88…(1)$
$p*2^2+q*2+81=289$
$4p+2q=208…(2)$
Solving (1) and (2), we get p =16, q =72, and r =81.



Therefore the general term formula is $16n^2+72n+81$. Now, if we start adding them up, 81 will be there 23 times. Then there will be 72, 144, 216 etc, 22 times. This sum will be 72(1+2+3+...22) =18216. At the last 16n^2 will be there 22 times. This sum will be $16(1^2+2^2+3^2...22^2) = 16 ((22) (23) (45)/6) = 60720$. Grand total is $1863+18216+60720-92= 80707$.






arithmetic-progressions







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$endgroup$












  • $begingroup$
    $15cdot 19=17^2-2^2$
    $endgroup$
    – giannispapav
    Dec 19 '18 at 14:34










  • $begingroup$
    The sequence ${9,13,17,21,ldots}$ is arithmetic with common difference $4$, so the $n$-th term is $a_n=4n+5$. Then the sum you want to evaluate is $sumlimits_{n=1}^{23}(4n+5-2)(4n+5+2)=sumlimits_{n=1}^{23}(16n^2+40n+21)$.
    $endgroup$
    – user170231
    Dec 19 '18 at 14:43


















0












$begingroup$


Solution: My attempt:
=$9^2-2^2 +13^2-2^2+17^2-2^2...97^2-2^2$

=$(9^2+13^2+17^2...97^2)-2^2( 23 )$
Focus on the first part. 81+169+289+441...Here first-order differences are 88, 120, 152... and the second-order difference is common to be 32. Therefore the series is of the form of $p(n)^2+q(n)+r$. If we plug n=0, we get the first term, if plug n=1, we get the second term, if we plug n=2, we get the third term and so on.
$p*0^2+q*0+r=81$
r=81
$p*1^2*q*1+81=169$
$p+q+81=169$
$p+q=88…(1)$
$p*2^2+q*2+81=289$
$4p+2q=208…(2)$
Solving (1) and (2), we get p =16, q =72, and r =81.



Therefore the general term formula is $16n^2+72n+81$. Now, if we start adding them up, 81 will be there 23 times. Then there will be 72, 144, 216 etc, 22 times. This sum will be 72(1+2+3+...22) =18216. At the last 16n^2 will be there 22 times. This sum will be $16(1^2+2^2+3^2...22^2) = 16 ((22) (23) (45)/6) = 60720$. Grand total is $1863+18216+60720-92= 80707$.






arithmetic-progressions







share|cite|improve this question











$endgroup$












  • $begingroup$
    $15cdot 19=17^2-2^2$
    $endgroup$
    – giannispapav
    Dec 19 '18 at 14:34










  • $begingroup$
    The sequence ${9,13,17,21,ldots}$ is arithmetic with common difference $4$, so the $n$-th term is $a_n=4n+5$. Then the sum you want to evaluate is $sumlimits_{n=1}^{23}(4n+5-2)(4n+5+2)=sumlimits_{n=1}^{23}(16n^2+40n+21)$.
    $endgroup$
    – user170231
    Dec 19 '18 at 14:43
















0












0








0


1



$begingroup$


Solution: My attempt:
=$9^2-2^2 +13^2-2^2+17^2-2^2...97^2-2^2$

=$(9^2+13^2+17^2...97^2)-2^2( 23 )$
Focus on the first part. 81+169+289+441...Here first-order differences are 88, 120, 152... and the second-order difference is common to be 32. Therefore the series is of the form of $p(n)^2+q(n)+r$. If we plug n=0, we get the first term, if plug n=1, we get the second term, if we plug n=2, we get the third term and so on.
$p*0^2+q*0+r=81$
r=81
$p*1^2*q*1+81=169$
$p+q+81=169$
$p+q=88…(1)$
$p*2^2+q*2+81=289$
$4p+2q=208…(2)$
Solving (1) and (2), we get p =16, q =72, and r =81.



Therefore the general term formula is $16n^2+72n+81$. Now, if we start adding them up, 81 will be there 23 times. Then there will be 72, 144, 216 etc, 22 times. This sum will be 72(1+2+3+...22) =18216. At the last 16n^2 will be there 22 times. This sum will be $16(1^2+2^2+3^2...22^2) = 16 ((22) (23) (45)/6) = 60720$. Grand total is $1863+18216+60720-92= 80707$.






arithmetic-progressions







share|cite|improve this question











$endgroup$




Solution: My attempt:
=$9^2-2^2 +13^2-2^2+17^2-2^2...97^2-2^2$

=$(9^2+13^2+17^2...97^2)-2^2( 23 )$
Focus on the first part. 81+169+289+441...Here first-order differences are 88, 120, 152... and the second-order difference is common to be 32. Therefore the series is of the form of $p(n)^2+q(n)+r$. If we plug n=0, we get the first term, if plug n=1, we get the second term, if we plug n=2, we get the third term and so on.
$p*0^2+q*0+r=81$
r=81
$p*1^2*q*1+81=169$
$p+q+81=169$
$p+q=88…(1)$
$p*2^2+q*2+81=289$
$4p+2q=208…(2)$
Solving (1) and (2), we get p =16, q =72, and r =81.



Therefore the general term formula is $16n^2+72n+81$. Now, if we start adding them up, 81 will be there 23 times. Then there will be 72, 144, 216 etc, 22 times. This sum will be 72(1+2+3+...22) =18216. At the last 16n^2 will be there 22 times. This sum will be $16(1^2+2^2+3^2...22^2) = 16 ((22) (23) (45)/6) = 60720$. Grand total is $1863+18216+60720-92= 80707$.






arithmetic-progressions




arithmetic-progressions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 19 '18 at 20:52









Lorenzo B.

1,8602520




1,8602520










asked Dec 19 '18 at 14:26









Sherlock WatsonSherlock Watson

3462413




3462413












  • $begingroup$
    $15cdot 19=17^2-2^2$
    $endgroup$
    – giannispapav
    Dec 19 '18 at 14:34










  • $begingroup$
    The sequence ${9,13,17,21,ldots}$ is arithmetic with common difference $4$, so the $n$-th term is $a_n=4n+5$. Then the sum you want to evaluate is $sumlimits_{n=1}^{23}(4n+5-2)(4n+5+2)=sumlimits_{n=1}^{23}(16n^2+40n+21)$.
    $endgroup$
    – user170231
    Dec 19 '18 at 14:43




















  • $begingroup$
    $15cdot 19=17^2-2^2$
    $endgroup$
    – giannispapav
    Dec 19 '18 at 14:34










  • $begingroup$
    The sequence ${9,13,17,21,ldots}$ is arithmetic with common difference $4$, so the $n$-th term is $a_n=4n+5$. Then the sum you want to evaluate is $sumlimits_{n=1}^{23}(4n+5-2)(4n+5+2)=sumlimits_{n=1}^{23}(16n^2+40n+21)$.
    $endgroup$
    – user170231
    Dec 19 '18 at 14:43


















$begingroup$
$15cdot 19=17^2-2^2$
$endgroup$
– giannispapav
Dec 19 '18 at 14:34




$begingroup$
$15cdot 19=17^2-2^2$
$endgroup$
– giannispapav
Dec 19 '18 at 14:34












$begingroup$
The sequence ${9,13,17,21,ldots}$ is arithmetic with common difference $4$, so the $n$-th term is $a_n=4n+5$. Then the sum you want to evaluate is $sumlimits_{n=1}^{23}(4n+5-2)(4n+5+2)=sumlimits_{n=1}^{23}(16n^2+40n+21)$.
$endgroup$
– user170231
Dec 19 '18 at 14:43






$begingroup$
The sequence ${9,13,17,21,ldots}$ is arithmetic with common difference $4$, so the $n$-th term is $a_n=4n+5$. Then the sum you want to evaluate is $sumlimits_{n=1}^{23}(4n+5-2)(4n+5+2)=sumlimits_{n=1}^{23}(16n^2+40n+21)$.
$endgroup$
– user170231
Dec 19 '18 at 14:43












2 Answers
2






active

oldest

votes


















1












$begingroup$

We have one more way to do it easily. As we can see the sequence has the general term $T_n$=$(4n+3)(4n+7)$ and the sum of sequence will be equal to $sum {T_n} $ from $n=1$ to $n=23$. Now, the sum will appear as :
$$16frac{n(n+1)(2n+1)}{6}+20n(n+1)+21n$$ where putting n=23 will give $80707$



It will be easier as there will be less chance for error as there will be less steps than what you have approached the question. Also what fellow users have given, it is just the same but easier coefficients to apply for.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks. How do you reach the general term? With induction?
    $endgroup$
    – Sherlock Watson
    Dec 19 '18 at 18:28










  • $begingroup$
    I used these points-(1)all are odd (2)all have the difference 4 (3)2nd part in the multiplication is also differing by 4 (4) So it must be (4k+r)(4k+q) and then put the value of 1,2,.(this doesn't mean putting all the values) to get r and 1 and then checked it through other numbers. I have used these steps in many questions (never fails me).
    $endgroup$
    – jayant98
    Dec 19 '18 at 18:40



















1












$begingroup$

begin{align}
S &= sum_{n=0}^{n=22} (4n+7)(4n+11)\
&= sum_{n=0}^{n=22} 16n^2+72n+color{red}{77}\
&= 80707
end{align}






share|cite|improve this answer











$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    We have one more way to do it easily. As we can see the sequence has the general term $T_n$=$(4n+3)(4n+7)$ and the sum of sequence will be equal to $sum {T_n} $ from $n=1$ to $n=23$. Now, the sum will appear as :
    $$16frac{n(n+1)(2n+1)}{6}+20n(n+1)+21n$$ where putting n=23 will give $80707$



    It will be easier as there will be less chance for error as there will be less steps than what you have approached the question. Also what fellow users have given, it is just the same but easier coefficients to apply for.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thanks. How do you reach the general term? With induction?
      $endgroup$
      – Sherlock Watson
      Dec 19 '18 at 18:28










    • $begingroup$
      I used these points-(1)all are odd (2)all have the difference 4 (3)2nd part in the multiplication is also differing by 4 (4) So it must be (4k+r)(4k+q) and then put the value of 1,2,.(this doesn't mean putting all the values) to get r and 1 and then checked it through other numbers. I have used these steps in many questions (never fails me).
      $endgroup$
      – jayant98
      Dec 19 '18 at 18:40
















    1












    $begingroup$

    We have one more way to do it easily. As we can see the sequence has the general term $T_n$=$(4n+3)(4n+7)$ and the sum of sequence will be equal to $sum {T_n} $ from $n=1$ to $n=23$. Now, the sum will appear as :
    $$16frac{n(n+1)(2n+1)}{6}+20n(n+1)+21n$$ where putting n=23 will give $80707$



    It will be easier as there will be less chance for error as there will be less steps than what you have approached the question. Also what fellow users have given, it is just the same but easier coefficients to apply for.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thanks. How do you reach the general term? With induction?
      $endgroup$
      – Sherlock Watson
      Dec 19 '18 at 18:28










    • $begingroup$
      I used these points-(1)all are odd (2)all have the difference 4 (3)2nd part in the multiplication is also differing by 4 (4) So it must be (4k+r)(4k+q) and then put the value of 1,2,.(this doesn't mean putting all the values) to get r and 1 and then checked it through other numbers. I have used these steps in many questions (never fails me).
      $endgroup$
      – jayant98
      Dec 19 '18 at 18:40














    1












    1








    1





    $begingroup$

    We have one more way to do it easily. As we can see the sequence has the general term $T_n$=$(4n+3)(4n+7)$ and the sum of sequence will be equal to $sum {T_n} $ from $n=1$ to $n=23$. Now, the sum will appear as :
    $$16frac{n(n+1)(2n+1)}{6}+20n(n+1)+21n$$ where putting n=23 will give $80707$



    It will be easier as there will be less chance for error as there will be less steps than what you have approached the question. Also what fellow users have given, it is just the same but easier coefficients to apply for.






    share|cite|improve this answer











    $endgroup$



    We have one more way to do it easily. As we can see the sequence has the general term $T_n$=$(4n+3)(4n+7)$ and the sum of sequence will be equal to $sum {T_n} $ from $n=1$ to $n=23$. Now, the sum will appear as :
    $$16frac{n(n+1)(2n+1)}{6}+20n(n+1)+21n$$ where putting n=23 will give $80707$



    It will be easier as there will be less chance for error as there will be less steps than what you have approached the question. Also what fellow users have given, it is just the same but easier coefficients to apply for.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 19 '18 at 14:51

























    answered Dec 19 '18 at 14:45









    jayant98jayant98

    657318




    657318












    • $begingroup$
      Thanks. How do you reach the general term? With induction?
      $endgroup$
      – Sherlock Watson
      Dec 19 '18 at 18:28










    • $begingroup$
      I used these points-(1)all are odd (2)all have the difference 4 (3)2nd part in the multiplication is also differing by 4 (4) So it must be (4k+r)(4k+q) and then put the value of 1,2,.(this doesn't mean putting all the values) to get r and 1 and then checked it through other numbers. I have used these steps in many questions (never fails me).
      $endgroup$
      – jayant98
      Dec 19 '18 at 18:40


















    • $begingroup$
      Thanks. How do you reach the general term? With induction?
      $endgroup$
      – Sherlock Watson
      Dec 19 '18 at 18:28










    • $begingroup$
      I used these points-(1)all are odd (2)all have the difference 4 (3)2nd part in the multiplication is also differing by 4 (4) So it must be (4k+r)(4k+q) and then put the value of 1,2,.(this doesn't mean putting all the values) to get r and 1 and then checked it through other numbers. I have used these steps in many questions (never fails me).
      $endgroup$
      – jayant98
      Dec 19 '18 at 18:40
















    $begingroup$
    Thanks. How do you reach the general term? With induction?
    $endgroup$
    – Sherlock Watson
    Dec 19 '18 at 18:28




    $begingroup$
    Thanks. How do you reach the general term? With induction?
    $endgroup$
    – Sherlock Watson
    Dec 19 '18 at 18:28












    $begingroup$
    I used these points-(1)all are odd (2)all have the difference 4 (3)2nd part in the multiplication is also differing by 4 (4) So it must be (4k+r)(4k+q) and then put the value of 1,2,.(this doesn't mean putting all the values) to get r and 1 and then checked it through other numbers. I have used these steps in many questions (never fails me).
    $endgroup$
    – jayant98
    Dec 19 '18 at 18:40




    $begingroup$
    I used these points-(1)all are odd (2)all have the difference 4 (3)2nd part in the multiplication is also differing by 4 (4) So it must be (4k+r)(4k+q) and then put the value of 1,2,.(this doesn't mean putting all the values) to get r and 1 and then checked it through other numbers. I have used these steps in many questions (never fails me).
    $endgroup$
    – jayant98
    Dec 19 '18 at 18:40











    1












    $begingroup$

    begin{align}
    S &= sum_{n=0}^{n=22} (4n+7)(4n+11)\
    &= sum_{n=0}^{n=22} 16n^2+72n+color{red}{77}\
    &= 80707
    end{align}






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      begin{align}
      S &= sum_{n=0}^{n=22} (4n+7)(4n+11)\
      &= sum_{n=0}^{n=22} 16n^2+72n+color{red}{77}\
      &= 80707
      end{align}






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        begin{align}
        S &= sum_{n=0}^{n=22} (4n+7)(4n+11)\
        &= sum_{n=0}^{n=22} 16n^2+72n+color{red}{77}\
        &= 80707
        end{align}






        share|cite|improve this answer











        $endgroup$



        begin{align}
        S &= sum_{n=0}^{n=22} (4n+7)(4n+11)\
        &= sum_{n=0}^{n=22} 16n^2+72n+color{red}{77}\
        &= 80707
        end{align}







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 19 '18 at 15:44

























        answered Dec 19 '18 at 14:40









        Karn WatcharasupatKarn Watcharasupat

        3,9742526




        3,9742526






























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            Dieringhausen

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            Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
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