show that these two equation holds by binomial theorem

Question

I know the binomial theorem, but I have no idea how to simplify this. I tried to write it as (y+x)^n+(y+x)^n-(y+x)^0+(y+x)^n-(y+x)^1+...+(y+x)^n-(y+x)^(n-1), but it didnt work out.

Answer 1

HINT:

For $k>0$ $$\frac{n!}{k!(n-k)!}k=n\frac{(n-1)!}{k\cdot (k-1)!\cdot \{n-1-(k-1)\}!}=n\binom{n-1}{k-1}$$

$$\frac{n!}{k!(n-k)!}k^2=\frac{n!}{k!(n-k)!}[k(k-1)+k]=k(k-1)\frac{n!}{k!(n-k)!}+k\frac{n!}{k!(n-k)!}$$

Now for $k>1$ $$k(k-1)\frac{n!}{k!(n-k)!}=k(k-1)n(n-1)\frac{(n-2)!}{k(k-1)\cdot (k-2)!\cdot[n-2-(k-2)]!}$$

$$=n(n-1)\binom{n-2}{k-2}$$

Answer 2

First problem: For fixed $y$, let $$f(x)=\sum_0^n \binom{n}{k}kx^k y^{n-k}.$$ Let $$g(x)=\sum_{k=0}^n \binom{n}{k}x^k y^{n-k}=(x+y)^n$$ Then $f(x)=xg'(x)=xn(x+y)^{n-1}$.

Second problem: Rewrite our sum as $$\sum_0^n \binom{n}{k}(k^2-k)x^ky^{n-k}+\sum_0^n \binom{n}{k}kx^ky^{n-k}.$$ The second sum has already been dealt with. The first sum is $x^2g''(x)$.