I need to simplify: $\sum_{k=1}^n ({n \choose k}k^2 )$. My first idea was to get rid of the square and use known identities to proceed. This is what I did
$$\sum_{k=1}^n( {n \choose k}k^2) = \sum_{k=1}^n( \frac n k {n-1 \choose k-1}k^2) = \sum_{k=1}^n {n-1 \choose k-1} nk = n \sum_{k=1}^n {n-1 \choose k-1} k = \\ n \sum_{s=0}^n \left({n-1 \choose s}(s+1) \right)$$
Now, the only thing that stops me here is the $s+1$ factor. Is there a way to dispose of it, or maybe to change it to $s$?
Asked
Active
Viewed 99 times
2

Aemilius
- 3,699
-
1Split it into two sums, one with the $s$ factor, the other with the $1$ factor. – Angina Seng Feb 24 '18 at 12:18
-
Differentiate $\sum_k \binom{n}{k}x^k$ – tired Feb 24 '18 at 12:19
-
i deleted my answer since it is redundant. See the contribution of @LeonBloy in the link given above – tired Feb 24 '18 at 13:00
2 Answers
2
Hint : $$\sum_{k=1}^n \binom{n}{k}k^2 =\sum_{k=1}^n \binom{n}{k}k(k-1)+ \sum_{k=1}^n \binom{n}{k}k$$ And if you write $$ P(x)= \sum_{k=1}^n \binom{n}{k}x^k= (1+x)^n -1$$ You will have : $$\sum_{k=1}^n \binom{n}{k}k^2 =P''(1)+P'(1)$$

Netchaiev
- 4,811
-
-
-
Yes. but I think that I already see it. I just had to take a closer look at this - it's simple. – Aemilius Feb 24 '18 at 12:38
1
Hint
You reached $$n \sum_{s=0}^{n-1} {n-1 \choose s}(s+1)$$ Let's continue $$n \sum_{s=0}^{n-1} {n-1 \choose s}(s+1) = n \left[\sum_{s=0}^{n-1} {n-1 \choose s}(s)+ \sum_{s=0}^{n-1} {n-1 \choose s}\right]$$ $$=n \left[(n-1)\sum_{s=1}^{n-1} {n-2 \choose {s-1}}+ \sum_{s=0}^{n-1} {n-1 \choose s} \right]$$ By binomial theorem
$$=n[(n-1)2^{n-2}+2^{n-1}]$$ $$=n(n+1).2^{n-2}$$

Rohan Shinde
- 9,737