Binomial Theorem Proof by Induction












5












$begingroup$


Did i prove the Binomial Theorem correctly? I got a feeling I did, but need another set of eyes to look over my work. Not really much of a question, sorry.



Binomial Theorem



$$(x+y)^{n}=sum_{k=0}^{n}{{n}choose{k}}x^{n-k}y^{k}$$



Base Case: $n=0$



$$(x+y)^{0}=1={{0}choose{0}}x^{0-0}y^{0}=sum_{k=0}^{0}{{0}choose{k}}x^{0-k}y^{k}$$



Induction Hypothesis



$$(x+y)^{n}=sum_{k=0}^{n}{{n}choose{k}}x^{n-k}y^{k}$$



Induction Step



$$begin{align*}
(x+y)^{n+1} &= (x+y)(x+y)^{n} \
&= xsum_{k=0}^{n}{{n}choose{k}}x^{n-k}y^{k}+ysum_{k=0}^{n}{{n}choose{k}}x^{n-k}y^{k} \
&= sum_{k=0}^{n}{{n}choose{k}}x^{n+1-k}y^{k}+sum_{k=0}^{n}{{n}choose{k}}x^{n-k}y^{k+1} \
&= {nchoose{0}}x^{n+1}+sum_{k=1}^{n}{{n}choose{k}}x^{n+1-k}y^{k}+{nchoose{n}}y^{n+1}+sum_{k=0}^{n-1}{{n}choose{k}}x^{n-k}y^{k+1} \
&= x^{n+1}+y^{n+1}+sum_{k=1}^{n}{{n}choose{k}}x^{n+1-k}y^{k}+sum_{k=0}^{n-1}{{n}choose{k}}x^{n-k}y^{k+1} \
&= {{n+1}choose{0}}x^{n+1}+{{n+1}choose{n+1}}y^{n+1}+sum_{k=1}^{n}{{n}choose{k}}x^{n+1-k}y^{k}+sum_{k=1}^{n}{{n}choose{k-1}}x^{n+1-k}y^{k} \
&= {{n+1}choose{0}}x^{n+1}+{{n+1}choose{n+1}}y^{n+1}+sum_{k=1}^{n}left({{n}choose{k}}+{{n}choose{k-1}}right)x^{n+1-k}y^{k} \
&= {{n+1}choose{0}}x^{n+1}+{{n+1}choose{n+1}}y^{n+1}+sum_{k=1}^{n}{{n+1}choose{k}}x^{n+1-k}y^{k} \
&= sum_{k=0}^{n+1}{{n+1}choose{k}}x^{n+1-k}y^{k}end{align*}$$






induction binomial-theorem







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  • 1




    $begingroup$
    Please write your work in mathjax here, rather than including only a picture. There are also several proofs of this here on MSE, on Wikipedia, and in many discrete math textbooks.
    $endgroup$
    – T. Bongers
    Mar 13 '16 at 6:16






  • 2




    $begingroup$
    Hard on the eyes to proofread handwritten text. But everything looks right, the key is reindexing so you can use the Pascal Identity, which you did without an explicit reference, $binom{n}{k}+binom{n}{k-1}=binom{n+1}{k}$.
    $endgroup$
    – André Nicolas
    Mar 13 '16 at 6:22


















5












$begingroup$


Did i prove the Binomial Theorem correctly? I got a feeling I did, but need another set of eyes to look over my work. Not really much of a question, sorry.



Binomial Theorem



$$(x+y)^{n}=sum_{k=0}^{n}{{n}choose{k}}x^{n-k}y^{k}$$



Base Case: $n=0$



$$(x+y)^{0}=1={{0}choose{0}}x^{0-0}y^{0}=sum_{k=0}^{0}{{0}choose{k}}x^{0-k}y^{k}$$



Induction Hypothesis



$$(x+y)^{n}=sum_{k=0}^{n}{{n}choose{k}}x^{n-k}y^{k}$$



Induction Step



$$begin{align*}
(x+y)^{n+1} &= (x+y)(x+y)^{n} \
&= xsum_{k=0}^{n}{{n}choose{k}}x^{n-k}y^{k}+ysum_{k=0}^{n}{{n}choose{k}}x^{n-k}y^{k} \
&= sum_{k=0}^{n}{{n}choose{k}}x^{n+1-k}y^{k}+sum_{k=0}^{n}{{n}choose{k}}x^{n-k}y^{k+1} \
&= {nchoose{0}}x^{n+1}+sum_{k=1}^{n}{{n}choose{k}}x^{n+1-k}y^{k}+{nchoose{n}}y^{n+1}+sum_{k=0}^{n-1}{{n}choose{k}}x^{n-k}y^{k+1} \
&= x^{n+1}+y^{n+1}+sum_{k=1}^{n}{{n}choose{k}}x^{n+1-k}y^{k}+sum_{k=0}^{n-1}{{n}choose{k}}x^{n-k}y^{k+1} \
&= {{n+1}choose{0}}x^{n+1}+{{n+1}choose{n+1}}y^{n+1}+sum_{k=1}^{n}{{n}choose{k}}x^{n+1-k}y^{k}+sum_{k=1}^{n}{{n}choose{k-1}}x^{n+1-k}y^{k} \
&= {{n+1}choose{0}}x^{n+1}+{{n+1}choose{n+1}}y^{n+1}+sum_{k=1}^{n}left({{n}choose{k}}+{{n}choose{k-1}}right)x^{n+1-k}y^{k} \
&= {{n+1}choose{0}}x^{n+1}+{{n+1}choose{n+1}}y^{n+1}+sum_{k=1}^{n}{{n+1}choose{k}}x^{n+1-k}y^{k} \
&= sum_{k=0}^{n+1}{{n+1}choose{k}}x^{n+1-k}y^{k}end{align*}$$






induction binomial-theorem







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Please write your work in mathjax here, rather than including only a picture. There are also several proofs of this here on MSE, on Wikipedia, and in many discrete math textbooks.
    $endgroup$
    – T. Bongers
    Mar 13 '16 at 6:16






  • 2




    $begingroup$
    Hard on the eyes to proofread handwritten text. But everything looks right, the key is reindexing so you can use the Pascal Identity, which you did without an explicit reference, $binom{n}{k}+binom{n}{k-1}=binom{n+1}{k}$.
    $endgroup$
    – André Nicolas
    Mar 13 '16 at 6:22
















5












5








5





$begingroup$


Did i prove the Binomial Theorem correctly? I got a feeling I did, but need another set of eyes to look over my work. Not really much of a question, sorry.



Binomial Theorem



$$(x+y)^{n}=sum_{k=0}^{n}{{n}choose{k}}x^{n-k}y^{k}$$



Base Case: $n=0$



$$(x+y)^{0}=1={{0}choose{0}}x^{0-0}y^{0}=sum_{k=0}^{0}{{0}choose{k}}x^{0-k}y^{k}$$



Induction Hypothesis



$$(x+y)^{n}=sum_{k=0}^{n}{{n}choose{k}}x^{n-k}y^{k}$$



Induction Step



$$begin{align*}
(x+y)^{n+1} &= (x+y)(x+y)^{n} \
&= xsum_{k=0}^{n}{{n}choose{k}}x^{n-k}y^{k}+ysum_{k=0}^{n}{{n}choose{k}}x^{n-k}y^{k} \
&= sum_{k=0}^{n}{{n}choose{k}}x^{n+1-k}y^{k}+sum_{k=0}^{n}{{n}choose{k}}x^{n-k}y^{k+1} \
&= {nchoose{0}}x^{n+1}+sum_{k=1}^{n}{{n}choose{k}}x^{n+1-k}y^{k}+{nchoose{n}}y^{n+1}+sum_{k=0}^{n-1}{{n}choose{k}}x^{n-k}y^{k+1} \
&= x^{n+1}+y^{n+1}+sum_{k=1}^{n}{{n}choose{k}}x^{n+1-k}y^{k}+sum_{k=0}^{n-1}{{n}choose{k}}x^{n-k}y^{k+1} \
&= {{n+1}choose{0}}x^{n+1}+{{n+1}choose{n+1}}y^{n+1}+sum_{k=1}^{n}{{n}choose{k}}x^{n+1-k}y^{k}+sum_{k=1}^{n}{{n}choose{k-1}}x^{n+1-k}y^{k} \
&= {{n+1}choose{0}}x^{n+1}+{{n+1}choose{n+1}}y^{n+1}+sum_{k=1}^{n}left({{n}choose{k}}+{{n}choose{k-1}}right)x^{n+1-k}y^{k} \
&= {{n+1}choose{0}}x^{n+1}+{{n+1}choose{n+1}}y^{n+1}+sum_{k=1}^{n}{{n+1}choose{k}}x^{n+1-k}y^{k} \
&= sum_{k=0}^{n+1}{{n+1}choose{k}}x^{n+1-k}y^{k}end{align*}$$






induction binomial-theorem







share|cite|improve this question











$endgroup$




Did i prove the Binomial Theorem correctly? I got a feeling I did, but need another set of eyes to look over my work. Not really much of a question, sorry.



Binomial Theorem



$$(x+y)^{n}=sum_{k=0}^{n}{{n}choose{k}}x^{n-k}y^{k}$$



Base Case: $n=0$



$$(x+y)^{0}=1={{0}choose{0}}x^{0-0}y^{0}=sum_{k=0}^{0}{{0}choose{k}}x^{0-k}y^{k}$$



Induction Hypothesis



$$(x+y)^{n}=sum_{k=0}^{n}{{n}choose{k}}x^{n-k}y^{k}$$



Induction Step



$$begin{align*}
(x+y)^{n+1} &= (x+y)(x+y)^{n} \
&= xsum_{k=0}^{n}{{n}choose{k}}x^{n-k}y^{k}+ysum_{k=0}^{n}{{n}choose{k}}x^{n-k}y^{k} \
&= sum_{k=0}^{n}{{n}choose{k}}x^{n+1-k}y^{k}+sum_{k=0}^{n}{{n}choose{k}}x^{n-k}y^{k+1} \
&= {nchoose{0}}x^{n+1}+sum_{k=1}^{n}{{n}choose{k}}x^{n+1-k}y^{k}+{nchoose{n}}y^{n+1}+sum_{k=0}^{n-1}{{n}choose{k}}x^{n-k}y^{k+1} \
&= x^{n+1}+y^{n+1}+sum_{k=1}^{n}{{n}choose{k}}x^{n+1-k}y^{k}+sum_{k=0}^{n-1}{{n}choose{k}}x^{n-k}y^{k+1} \
&= {{n+1}choose{0}}x^{n+1}+{{n+1}choose{n+1}}y^{n+1}+sum_{k=1}^{n}{{n}choose{k}}x^{n+1-k}y^{k}+sum_{k=1}^{n}{{n}choose{k-1}}x^{n+1-k}y^{k} \
&= {{n+1}choose{0}}x^{n+1}+{{n+1}choose{n+1}}y^{n+1}+sum_{k=1}^{n}left({{n}choose{k}}+{{n}choose{k-1}}right)x^{n+1-k}y^{k} \
&= {{n+1}choose{0}}x^{n+1}+{{n+1}choose{n+1}}y^{n+1}+sum_{k=1}^{n}{{n+1}choose{k}}x^{n+1-k}y^{k} \
&= sum_{k=0}^{n+1}{{n+1}choose{k}}x^{n+1-k}y^{k}end{align*}$$






induction binomial-theorem




induction binomial-theorem






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 23 '18 at 10:11









Sik Feng Cheong

1579




1579










asked Mar 13 '16 at 5:56









EdtheBigEdtheBig

46114




46114








  • 1




    $begingroup$
    Please write your work in mathjax here, rather than including only a picture. There are also several proofs of this here on MSE, on Wikipedia, and in many discrete math textbooks.
    $endgroup$
    – T. Bongers
    Mar 13 '16 at 6:16






  • 2




    $begingroup$
    Hard on the eyes to proofread handwritten text. But everything looks right, the key is reindexing so you can use the Pascal Identity, which you did without an explicit reference, $binom{n}{k}+binom{n}{k-1}=binom{n+1}{k}$.
    $endgroup$
    – André Nicolas
    Mar 13 '16 at 6:22
















  • 1




    $begingroup$
    Please write your work in mathjax here, rather than including only a picture. There are also several proofs of this here on MSE, on Wikipedia, and in many discrete math textbooks.
    $endgroup$
    – T. Bongers
    Mar 13 '16 at 6:16






  • 2




    $begingroup$
    Hard on the eyes to proofread handwritten text. But everything looks right, the key is reindexing so you can use the Pascal Identity, which you did without an explicit reference, $binom{n}{k}+binom{n}{k-1}=binom{n+1}{k}$.
    $endgroup$
    – André Nicolas
    Mar 13 '16 at 6:22










1




1




$begingroup$
Please write your work in mathjax here, rather than including only a picture. There are also several proofs of this here on MSE, on Wikipedia, and in many discrete math textbooks.
$endgroup$
– T. Bongers
Mar 13 '16 at 6:16




$begingroup$
Please write your work in mathjax here, rather than including only a picture. There are also several proofs of this here on MSE, on Wikipedia, and in many discrete math textbooks.
$endgroup$
– T. Bongers
Mar 13 '16 at 6:16




2




2




$begingroup$
Hard on the eyes to proofread handwritten text. But everything looks right, the key is reindexing so you can use the Pascal Identity, which you did without an explicit reference, $binom{n}{k}+binom{n}{k-1}=binom{n+1}{k}$.
$endgroup$
– André Nicolas
Mar 13 '16 at 6:22






$begingroup$
Hard on the eyes to proofread handwritten text. But everything looks right, the key is reindexing so you can use the Pascal Identity, which you did without an explicit reference, $binom{n}{k}+binom{n}{k-1}=binom{n+1}{k}$.
$endgroup$
– André Nicolas
Mar 13 '16 at 6:22












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