Evaluate $\sum_{k=1}^{n} (2k-1) {n \choose k} $ using calculus
I found out the value by the following method:
$$T_r= (2r-1) {n\choose r}$$
$$S_r= \sum^n _{r=1} \left(2r {n\choose r} – {n\choose r}\right) = 2n \sum^n_{r=1} {n-1\choose r-1} – \sum ^n _{r=1} {n\choose r}= 2n(2^{n-1}) -2^n +1$$
Can anyone help me find the same result using calculus*?
*by using the binomial expansion of : $(1-x)^n$
By using calculus I mean something like this:
For finding $1{n \choose 1} + 2{n\choose 2} + 3{n\choose 3} +…. + n{n \choose n}$ We can differentiate the expansion of $(1+x)^n$ with respect to $x$.
Best Answer
Consider the polynomial $$f(x)=\frac{(1+x^2)^n-1}{x}.$$ By the Binomial Theorem, for $1\le k\le n$, the coefficient of $x^{2k}$ in $(1+x^2)^n-1$ is $\binom{n}{k}$, and therefore the coefficient of $x^{2k-1}$ in $f(x)$ is $\binom{n}{k}$.
Differentiate $\sum_1^n \binom{n}{k}x^{2k-1}$ term by term and set $x=1$. We get our sum.
But note that the derivative of $f(x)$ is $$\frac{2nx(1+x^2)^{n-1}-((1+x^2)^n-1)}{x^2}.$$ Setting $x=1$ gives $2n\cdot 2^{n-1}-2^n+1$.