Solving for the finite sum of squares using calculus

[zoki85]

Is there some way that using calculus we can explicitely sum up :
\sum_{i=0}^{n}(i+1)^2 \binom{n}{i}=
?

For example,applying the derivative to the both sides of expression
1+x+x^2+...+x^n=\frac{x^{n+1}-1}{x-1}
we quickly get how much is 1+2x+3x^2+...+nx^{n-1}=.
But I don't know what to do with the first sum.Tryed binomial theorem and some identies but couldn't make use of them.
Thanks for your advice in advance.

 

[nworm]

You can use this ideas:

(i+1)^2=i^2+2i+1

.
i^2\times\frac{n!}{i!(n-i)!}=i\times\frac{i\times n!}{i!(n-i)!}=
.

i\times\frac{n\times (n-1)!}{(i-1)!(n-i)!}=<br /> i\times\frac{n\times (n-1)!}{(i-1)!(n-i)!}=
.

n\times\frac{(i-1+1)\times (n-1)!}{(i-1)!(n-i)!}=
.
n\times\frac{(n-1)!}{(i-1)!(n-i)!}+n\times\frac{(i-1)\times (n-1)!}{(i-1)!(n-i)!}=

.
n\times\frac{(n-1)!}{(i-1)!(n-i)!}+n\times\frac{(n-1)\times (n-2)!}{(i-2)!(n-i)!}

 

[zoki85]

Well I tryed and I couldn't make much of progress with this.
This problem I found in one of my calculus textbooks and I think we are supposed to use calculus methods somehow...

 

[yip]

I think you can do the sum like this

(1+x)^n=nC0+nC1x+...+nCnx^n
Multiply both sides by x
x(1+x)^n=nC0x+nC1x^2+...+nCnx^(n+1)
Differentiate both sides wrtx
(1+x)^n+nx(1+x)^(n-1)=nC0+nC12x+...+nCn(n+1)x^n
Multiply both sides by x
x[(1+x)^n+nx(1+x)^(n-1)]=nC0x+nC12x^2+...+nCn(n+1)x^(n+1)
Differentiate both sides wrtx
(1+x)^n+nx(1+x)^(n-1)+2nx(1+x)^(n-1)+n(n-1)(x^2)(1+x)^(n-2)=nC0x+nC12^2x+...+nCn}(n+1)^2x^n
Let x=1, after some simplification,
Sum=[2^(n-2)](n+4)(n+1)

 

[zoki85]

Thanks!I knew of the result but didn't know how to derive it.

 

[zoki85]

Your solution gave me idea to find another solution with calculus method.
Basically the two are equivavalent but I played with antiderivatives.

If f(x)=1^2 + 2^2\binom{n}{1}x + 3^2\binom{n}{2}x^2 + ...+(n+1)^2\binom{n}{n}x^n

Then F(x)=\int_{0}^{x}f(t)dt=x+2\binom{n}{1}x +3\binom{n}{2}x^2+...+(n+1)\binom{n}{n}x^n

g(x)=\frac{F(x)}{x}=1+2\binom{n}{1}x + 3\binom{n}{2}x^2+...+(n+1)\binom{n}{n}x^n

Integrating one more time:
G(x)=\int_{0}^{x}g(t)dt=x+\binom{n}{1}x^2+...+\binom{n}{n}x^{n+1}=x(1+x)^n

Now we conclude:
g(x)=(1+x)^n+nx(1+x)^{n-1}
F(x)=x(1+x)^n +nx^2(1+x)^{n-1}
f(x)=(1+x)^n +3nx(1+x)^{n-1}+n(n-1)x^2(1+x)^{n-2}

Substitution x=1 in the last one gives the result.