# Determine: $S = \frac{1}{2}{n \choose 0} + \frac{1}{3}{n \choose 1} + \cdots + \frac{1}{n+2}{n \choose n}$ [duplicate]

I am studying the book Equations and Inequalities by Herman et al, and am stuck on the following exercise:

Determine: $S = \frac{1}{2}{n \choose 0} + \frac{1}{3}{n \choose 1} + \cdots + \frac{1}{n+2}{n \choose n}$

The hint says to consider $(n+2)(n+1)S$.

Doing so, I get:

$(n+2)(n+1)S = \frac{1}{2}{n+2 \choose 0} + \frac{1}{3}{n+2 \choose 1} + \cdots + \frac{1}{n+2}{n+2 \choose n}$,

but I don't see how this is any easier to solve.

I wonder if somebody might help by expanding on the given hint, and maybe offering their suggestions as to how to use it properly.

Thanks.

• $(n+1)(n+2)\binom nk> \frac{(n+1)(n+2)}{(n-k+1)(n-k+2)}\binom nk=\binom{n+2}k$. – ajotatxe Mar 4 '17 at 17:48
• – StubbornAtom Mar 4 '17 at 18:12

Integration Approach

Since \begin{align} \frac{\mathrm{d}}{\mathrm{d}x}\sum_{k=0}^n\frac1{k+2}\binom{n}{k}x^{k+2} &=\sum_{k=0}^n\binom{n}{k}x^{k+1}\\ &=x(1+x)^n \end{align} we have \begin{align} \sum_{k=0}^n\frac1{k+2}\binom{n}{k} &=\int_0^1x(1+x)^n\,\mathrm{d}x\\ &=\int_1^2(x-1)x^n\,\mathrm{d}x\\ &=\frac{2^{n+2}-1}{n+2}-\frac{2^{n+1}-1}{n+1} \end{align}

Pre-calculus Approach

lab bhattacharjee has already given a hint for this approach, but I was working on adding it so I will include it.

Since $$\binom{n}{k}=\frac{(k+2)(k+1)}{(n+2)(n+1)}\binom{n+2}{k+2}$$ and $$\binom{n+1}{k+1}=\frac{k+2}{n+2}\binom{n+2}{k+2}$$ we have \begin{align} \sum_{k=0}^n\frac1{k+2}\binom{n}{k} &=\frac1{(n+1)(n+2)}\sum_{k=0}^n(k+1)\binom{n+2}{k+2}\\ &=\frac1{(n+1)(n+2)}\left[\sum_{k=0}^n(k+2)\binom{n+2}{k+2}-\sum_{k=0}^n\binom{n+2}{k+2}\right]\\ &=\frac1{(n+1)(n+2)}\left[(n+2)\sum_{k=0}^n\binom{n+1}{k+1}-\sum_{k=0}^n\binom{n+2}{k+2}\right]\\ &=\frac1{(n+1)(n+2)}\left[(n+2)\left(2^{n+1}-1\right)-\left(2^{n+2}-(n+2)-1\right)\right]\\ &=\frac{2^{n+1}}{n+1}-\frac{2^{n+2}-1}{(n+1)(n+2)} \end{align}

Noting that $\frac1{(n+1)(n+2)}=\frac1{n+1}-\frac1{n+2}$ and $2^{n+2}=2\cdot2^{n+1}$, we see that the two approaches give the same answer.

• This answer is the most explicit, and I believe has taught me the most. Thank you. – Adam Mar 4 '17 at 20:41

$$\dfrac1{k+2}\binom nk=\dfrac{k+1}{(n+1)(n+2)}\cdot\dfrac{(n+2)!}{(k+2)!\{n+2-(k+2)\}!}=\dfrac{k+1}{(n+1)(n+2)}\cdot\binom{n+2}{k+2}$$

Now $\displaystyle(k+1)\cdot\binom{n+2}{k+2}$

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

$\displaystyle=(n+2)\cdot\binom{(n+1)!}{(k+1)!\{n+1-(k+1)\}!}-\binom{n+2}{k+2}$

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

Now use $\displaystyle(1+1)^m=\sum_{r=0}^m\binom mr$

• (+1) This is the answer that I would usually have gone for, but I posted an integration answer for a change. One still has to be careful about the limits of summation in this approach. – robjohn Mar 4 '17 at 18:17
• Thanks. This looks more like the answer I would have been expected to produce given the techniques presented in the book. But I can see that the integration one also gives the right result. – Adam Mar 4 '17 at 18:20

Hint : Consider the function $$f(x) = \sum_{k=2}^{n+2}\frac{x^k}{k}\binom{n+2}{k}.$$ Then, compute $f'(x)$ and express it in a closed form, using the binomial expansion theorem. The value you wanna find is $f(1)$.