Is this proof for $sum^n_{i=1} i= frac{n^2+n}{2}$ correct?












3












$begingroup$


Is this proof correct, as I feel unsure about whether or not I did that correct because the book did it differently, I wouldn't know however why my proof should be wrong.



Could you help me out?



The following statement is to be proven by induction.
$$sum^n_{i=1} = frac{n^2+n}{2}$$
Base case $n=1$
$$1 = frac{1+1}{2} checkmark $$
Induction Step $nrightarrow n+1$
$$sum^{n+1}_{i=1}=sum^n_{i=1}+(n+1)\
iff frac{n^2+n}{2}+(n+1) \
iff frac{n^2+n}{2}+frac{2(n+1)}{2} \
iff frac{n^2+n+2n+2}{2}\
iff frac{n^2+3n+2}{2} \
iff frac{(n+1)^2+(n+1)}{2} $$






proof-verification







share|cite|improve this question











$endgroup$












  • $begingroup$
    Your proof is correct.
    $endgroup$
    – Jochen
    Dec 17 '18 at 7:47










  • $begingroup$
    Let try to improve you question using MathJax.
    $endgroup$
    – gimusi
    Dec 17 '18 at 7:50










  • $begingroup$
    @thebilly have you seen Gauss' solution to this problem, it is quite elegant.
    $endgroup$
    – Mustafa Said
    Dec 17 '18 at 23:34










  • $begingroup$
    @gimusi Yes I could do that, but that is a lot of work... I'd spend more time writing this into mathjax than solving my problems, which i guess is rather bad time management in regards to exams. is it not ok to post quick questions in pictures?
    $endgroup$
    – thebilly
    Dec 18 '18 at 20:01








  • 1




    $begingroup$
    @gimusi i see. I'll edit it.
    $endgroup$
    – thebilly
    Dec 18 '18 at 21:39
















3












$begingroup$


Is this proof correct, as I feel unsure about whether or not I did that correct because the book did it differently, I wouldn't know however why my proof should be wrong.



Could you help me out?



The following statement is to be proven by induction.
$$sum^n_{i=1} = frac{n^2+n}{2}$$
Base case $n=1$
$$1 = frac{1+1}{2} checkmark $$
Induction Step $nrightarrow n+1$
$$sum^{n+1}_{i=1}=sum^n_{i=1}+(n+1)\
iff frac{n^2+n}{2}+(n+1) \
iff frac{n^2+n}{2}+frac{2(n+1)}{2} \
iff frac{n^2+n+2n+2}{2}\
iff frac{n^2+3n+2}{2} \
iff frac{(n+1)^2+(n+1)}{2} $$






proof-verification







share|cite|improve this question











$endgroup$












  • $begingroup$
    Your proof is correct.
    $endgroup$
    – Jochen
    Dec 17 '18 at 7:47










  • $begingroup$
    Let try to improve you question using MathJax.
    $endgroup$
    – gimusi
    Dec 17 '18 at 7:50










  • $begingroup$
    @thebilly have you seen Gauss' solution to this problem, it is quite elegant.
    $endgroup$
    – Mustafa Said
    Dec 17 '18 at 23:34










  • $begingroup$
    @gimusi Yes I could do that, but that is a lot of work... I'd spend more time writing this into mathjax than solving my problems, which i guess is rather bad time management in regards to exams. is it not ok to post quick questions in pictures?
    $endgroup$
    – thebilly
    Dec 18 '18 at 20:01








  • 1




    $begingroup$
    @gimusi i see. I'll edit it.
    $endgroup$
    – thebilly
    Dec 18 '18 at 21:39














3












3








3





$begingroup$


Is this proof correct, as I feel unsure about whether or not I did that correct because the book did it differently, I wouldn't know however why my proof should be wrong.



Could you help me out?



The following statement is to be proven by induction.
$$sum^n_{i=1} = frac{n^2+n}{2}$$
Base case $n=1$
$$1 = frac{1+1}{2} checkmark $$
Induction Step $nrightarrow n+1$
$$sum^{n+1}_{i=1}=sum^n_{i=1}+(n+1)\
iff frac{n^2+n}{2}+(n+1) \
iff frac{n^2+n}{2}+frac{2(n+1)}{2} \
iff frac{n^2+n+2n+2}{2}\
iff frac{n^2+3n+2}{2} \
iff frac{(n+1)^2+(n+1)}{2} $$






proof-verification







share|cite|improve this question











$endgroup$




Is this proof correct, as I feel unsure about whether or not I did that correct because the book did it differently, I wouldn't know however why my proof should be wrong.



Could you help me out?



The following statement is to be proven by induction.
$$sum^n_{i=1} = frac{n^2+n}{2}$$
Base case $n=1$
$$1 = frac{1+1}{2} checkmark $$
Induction Step $nrightarrow n+1$
$$sum^{n+1}_{i=1}=sum^n_{i=1}+(n+1)\
iff frac{n^2+n}{2}+(n+1) \
iff frac{n^2+n}{2}+frac{2(n+1)}{2} \
iff frac{n^2+n+2n+2}{2}\
iff frac{n^2+3n+2}{2} \
iff frac{(n+1)^2+(n+1)}{2} $$






proof-verification




proof-verification






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 18 '18 at 21:46







thebilly

















asked Dec 17 '18 at 7:45









thebillythebilly

566




566












  • $begingroup$
    Your proof is correct.
    $endgroup$
    – Jochen
    Dec 17 '18 at 7:47










  • $begingroup$
    Let try to improve you question using MathJax.
    $endgroup$
    – gimusi
    Dec 17 '18 at 7:50










  • $begingroup$
    @thebilly have you seen Gauss' solution to this problem, it is quite elegant.
    $endgroup$
    – Mustafa Said
    Dec 17 '18 at 23:34










  • $begingroup$
    @gimusi Yes I could do that, but that is a lot of work... I'd spend more time writing this into mathjax than solving my problems, which i guess is rather bad time management in regards to exams. is it not ok to post quick questions in pictures?
    $endgroup$
    – thebilly
    Dec 18 '18 at 20:01








  • 1




    $begingroup$
    @gimusi i see. I'll edit it.
    $endgroup$
    – thebilly
    Dec 18 '18 at 21:39


















  • $begingroup$
    Your proof is correct.
    $endgroup$
    – Jochen
    Dec 17 '18 at 7:47










  • $begingroup$
    Let try to improve you question using MathJax.
    $endgroup$
    – gimusi
    Dec 17 '18 at 7:50










  • $begingroup$
    @thebilly have you seen Gauss' solution to this problem, it is quite elegant.
    $endgroup$
    – Mustafa Said
    Dec 17 '18 at 23:34










  • $begingroup$
    @gimusi Yes I could do that, but that is a lot of work... I'd spend more time writing this into mathjax than solving my problems, which i guess is rather bad time management in regards to exams. is it not ok to post quick questions in pictures?
    $endgroup$
    – thebilly
    Dec 18 '18 at 20:01








  • 1




    $begingroup$
    @gimusi i see. I'll edit it.
    $endgroup$
    – thebilly
    Dec 18 '18 at 21:39
















$begingroup$
Your proof is correct.
$endgroup$
– Jochen
Dec 17 '18 at 7:47




$begingroup$
Your proof is correct.
$endgroup$
– Jochen
Dec 17 '18 at 7:47












$begingroup$
Let try to improve you question using MathJax.
$endgroup$
– gimusi
Dec 17 '18 at 7:50




$begingroup$
Let try to improve you question using MathJax.
$endgroup$
– gimusi
Dec 17 '18 at 7:50












$begingroup$
@thebilly have you seen Gauss' solution to this problem, it is quite elegant.
$endgroup$
– Mustafa Said
Dec 17 '18 at 23:34




$begingroup$
@thebilly have you seen Gauss' solution to this problem, it is quite elegant.
$endgroup$
– Mustafa Said
Dec 17 '18 at 23:34












$begingroup$
@gimusi Yes I could do that, but that is a lot of work... I'd spend more time writing this into mathjax than solving my problems, which i guess is rather bad time management in regards to exams. is it not ok to post quick questions in pictures?
$endgroup$
– thebilly
Dec 18 '18 at 20:01






$begingroup$
@gimusi Yes I could do that, but that is a lot of work... I'd spend more time writing this into mathjax than solving my problems, which i guess is rather bad time management in regards to exams. is it not ok to post quick questions in pictures?
$endgroup$
– thebilly
Dec 18 '18 at 20:01






1




1




$begingroup$
@gimusi i see. I'll edit it.
$endgroup$
– thebilly
Dec 18 '18 at 21:39




$begingroup$
@gimusi i see. I'll edit it.
$endgroup$
– thebilly
Dec 18 '18 at 21:39










1 Answer
1






active

oldest

votes


















0












$begingroup$

Yes it is correct indeed, also according to the usual way to write that foundamental identity, that is



$$sum^n_{i=1} i= frac{n(n+1)}{2}$$



at the end we obtain



$$sum^{n+1}_{i=1} i= frac{n^2+3n+2}{2}= frac{(n+1)(n+2)}{2}$$



Refer also to the related




  • Using Direct Proof. $1+2+3+ldots+n = frac{n(n + 1)}{2}$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I see. Thank you.
    $endgroup$
    – thebilly
    Dec 18 '18 at 20:02










  • $begingroup$
    @thebilly You are welcome bye.
    $endgroup$
    – gimusi
    Dec 18 '18 at 21:14











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

Yes it is correct indeed, also according to the usual way to write that foundamental identity, that is



$$sum^n_{i=1} i= frac{n(n+1)}{2}$$



at the end we obtain



$$sum^{n+1}_{i=1} i= frac{n^2+3n+2}{2}= frac{(n+1)(n+2)}{2}$$



Refer also to the related




  • Using Direct Proof. $1+2+3+ldots+n = frac{n(n + 1)}{2}$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I see. Thank you.
    $endgroup$
    – thebilly
    Dec 18 '18 at 20:02










  • $begingroup$
    @thebilly You are welcome bye.
    $endgroup$
    – gimusi
    Dec 18 '18 at 21:14
















0












$begingroup$

Yes it is correct indeed, also according to the usual way to write that foundamental identity, that is



$$sum^n_{i=1} i= frac{n(n+1)}{2}$$



at the end we obtain



$$sum^{n+1}_{i=1} i= frac{n^2+3n+2}{2}= frac{(n+1)(n+2)}{2}$$



Refer also to the related




  • Using Direct Proof. $1+2+3+ldots+n = frac{n(n + 1)}{2}$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I see. Thank you.
    $endgroup$
    – thebilly
    Dec 18 '18 at 20:02










  • $begingroup$
    @thebilly You are welcome bye.
    $endgroup$
    – gimusi
    Dec 18 '18 at 21:14














0












0








0





$begingroup$

Yes it is correct indeed, also according to the usual way to write that foundamental identity, that is



$$sum^n_{i=1} i= frac{n(n+1)}{2}$$



at the end we obtain



$$sum^{n+1}_{i=1} i= frac{n^2+3n+2}{2}= frac{(n+1)(n+2)}{2}$$



Refer also to the related




  • Using Direct Proof. $1+2+3+ldots+n = frac{n(n + 1)}{2}$






share|cite|improve this answer











$endgroup$



Yes it is correct indeed, also according to the usual way to write that foundamental identity, that is



$$sum^n_{i=1} i= frac{n(n+1)}{2}$$



at the end we obtain



$$sum^{n+1}_{i=1} i= frac{n^2+3n+2}{2}= frac{(n+1)(n+2)}{2}$$



Refer also to the related




  • Using Direct Proof. $1+2+3+ldots+n = frac{n(n + 1)}{2}$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 17 '18 at 23:15

























answered Dec 17 '18 at 7:48









gimusigimusi

92.8k84494




92.8k84494












  • $begingroup$
    I see. Thank you.
    $endgroup$
    – thebilly
    Dec 18 '18 at 20:02










  • $begingroup$
    @thebilly You are welcome bye.
    $endgroup$
    – gimusi
    Dec 18 '18 at 21:14


















  • $begingroup$
    I see. Thank you.
    $endgroup$
    – thebilly
    Dec 18 '18 at 20:02










  • $begingroup$
    @thebilly You are welcome bye.
    $endgroup$
    – gimusi
    Dec 18 '18 at 21:14
















$begingroup$
I see. Thank you.
$endgroup$
– thebilly
Dec 18 '18 at 20:02




$begingroup$
I see. Thank you.
$endgroup$
– thebilly
Dec 18 '18 at 20:02












$begingroup$
@thebilly You are welcome bye.
$endgroup$
– gimusi
Dec 18 '18 at 21:14




$begingroup$
@thebilly You are welcome bye.
$endgroup$
– gimusi
Dec 18 '18 at 21:14


















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