Given question :$C_0^2-2C_1^2+3C_2^2-....+(-1)^n(n+1)C_n^2$
now i divided the question in two parts
- I founded the sum of seires by comparing coefficient
$C_0^2-C_1^2+ C_2^2-....+(-1)^nC_n^2$
$[(1-x)(1+x)]^n$=$\frac{(−1)^{n/2}n!}{(n/2)!(n/2)}$
2. The series left
$-1C_1^2+2C_2^2-....+(-1)^n(n)C_n^2$
first i applied the formula $r*^{n}C_r=n*^{n-1}C_r-1$
n*[$^{n-1}C_0*^{n}C_1-^{n-1}C_1*^{n}C_2...$]
but now due to alternate plus minus sign i am not able to proceed as comparing the coefficient method does not work as base and power are different - $(1-x)^{n-1}(1+x)^n$ how can we i further proceed this question?
Hint:
$$(r+1)\binom nr=\binom nr+r\binom nr=\cdots=\binom nr+n\binom{n-1}{r-1}$$
So, we need $$\sum_{r=0}^n(-1)^r(r+1)\binom nr\cdot\binom nr=\sum_{r=0}^n(-1)^r\binom nr\cdot\binom nr+n\sum_{r=0}^n(-1)^r\binom nr\cdot\binom{n-1}{r-1}$$
Now for the second sum, the term containing $x^n$ in
$$(1-x)^n(x+1)^{n-1}=(1-x)(1-x^2)^{n-1}$$
is $$\binom nr(-x)^r\binom{n-1}{r-1}x^{n-r}=x^n\binom nr\binom{n-1}{r-1}(-1)^r$$
So, if $n$ is odd the coefficient of $x^{2m+1}$ in $$(1-x)(1-x^2)^{2m}$$ will be $\displaystyle -(-1)^m\binom{2m}m$
What if $n$ is even $=2m$(say)