Prove that \[1{n\choose 1} + 2{n\choose 2} + 3{n \choose 3} + \cdots + n{n\choose n} = n2^{n-1}\] for all positive integers n.
This might be a repost of one of Mellie's questions, but I don't know.
Here is a "proof" or verification for 3 terms only but it holds true for any n terms:
Expand the following:
binomial(n, 1) + 2 binomial(n, 2) + 3 binomial(n, 3) = n×2^(n - 1)
n (n - 1) = n n - n:
binomial(n, 1) + n n - n + 3 binomial(n, 3) = n×2^(n - 1)
n n = n^2:
binomial(n, 1) - n + n^2 + 3 binomial(n, 3) = n×2^(n - 1)
(n - 2) (n - 1) = (n) (n) + (n) (-1) + (-2) (n) + (-2) (-1) = n^2 - n - 2 n + 2 = n^2 - 3 n + 2:
binomial(n, 1) - n + n^2 + (n^2 - 3 n + 2 n)/2 = n×2^(n - 1)
n/2 (n^2 - 3 n + 2) = (n n^2)/2 + 1/2 n (-3 n) + (n 2)/2:
(n n^2)/2 - (3 n n)/2 + (2 n)/2 + n^2 + binomial(n, 1) - n = n×2^(n - 1)
(n (-3) n)/2 = (-3 n^2)/2:
(n n^2)/2 + n^2 + (-3 n^2)/2 + (2 n)/2 + binomial(n, 1) - n = n×2^(n - 1)
(n n^2)/2 = n^(1 + 2)/2:
n^(1 + 2)/2 + n^2 - (3 n^2)/2 + (2 n)/2 + binomial(n, 1) - n = n×2^(n - 1)
1 + 2 = 3:
n^3/2 + n^2 - (3 n^2)/2 + (2 n)/2 + binomial(n, 1) - n = n×2^(n - 1)
Grouping like terms, n^3/2 + n^2 + 1/2 n^2 (-3) + (n 2)/2 + binomial(n, 1) + n (-1) = n^3/2 + (n^2 - (3 n^2)/2) + (n - n + n):
n^3/2 + (n^2 - (3 n^2)/2) + (n - n + n) = n×2^(n - 1)
n^2 - (3 n^2)/2 = -n^2/2:
n^3/2 + -(n^2)/2 + (n - n + n) = n×2^(n - 1)
n - n + n = n:
n^3/2 - n^2/2 + n = n×2^(n - 1) Courtesy of Mathematica 11 Home Edition.
+26367 Prove that \[1{n\choose 1} + 2{n\choose 2} + 3{n \choose 3} + \cdots + n{n\choose n} = n2^{n-1}\]
for all positive integers n.
\(\begin{array}{|rcll|} \hline && \mathbf{1\dbinom n1 + 2\dbinom n2 + 3\dbinom n3 + 4\dbinom n4 + \ldots + n\dbinom nn} \\\\ &=& 1\cdot\dfrac{n!}{(n-1)!\cdot 1} + 2\cdot\dfrac{n!}{(n-2)!\cdot 1\cdot 2} + 3\cdot\dfrac{n!}{(n-3)!\cdot 1\cdot 2\cdot 3} \\ &&+ 4\cdot\dfrac{n!}{(n-4)!\cdot 1\cdot 2\cdot 3\cdot 4} + \ldots + n\cdot\dfrac{n!}{(n-n)!\cdot 1\cdot 2\cdot \ldots \cdot(n-1)\cdot n } \\\\ &=& \dfrac{n!}{(n-1)!\cdot 0!} + \dfrac{n!}{(n-2)!\cdot 1!} + \dfrac{n!}{(n-3)!\cdot 2!} + \dfrac{n!}{(n-4)!\cdot 3!} \\ && +\ldots + \dfrac{n!}{(n-n)!\cdot(n-1)! } \\\\ &&\boxed{n!=n\cdot(n-1)!} \\\\ &=& \dfrac{n\cdot(n-1)!}{(n-1)!\cdot 0!} + \dfrac{n\cdot(n-1)!}{(n-2)!\cdot 1!} + \dfrac{n\cdot(n-1)!}{(n-3)!\cdot 2!} + \dfrac{n\cdot(n-1)!}{(n-4)!\cdot 3!} \\ && +\ldots + \dfrac{n\cdot(n-1)!}{(n-n)!\cdot(n-1)! } \\\\ &=& n\cdot \Big[ \dfrac{ (n-1)!}{(n-1)!\cdot 0!} + \dfrac{ (n-1)!}{(n-2)!\cdot 1!} + \dfrac{ (n-1)!}{(n-3)!\cdot 2!} + \dfrac{ (n-1)!}{(n-4)!\cdot 3!} \\ && +\ldots + \dfrac{ (n-1)!}{(n-n)!\cdot(n-1)! } \Big] \\\\ &=& n\cdot \Big[ \underbrace{\dbinom{n-1}{0}+\dbinom{n-1}{1}+\dbinom{n-1}{2}+\dbinom{n-1}{3} +\ldots + \dbinom{n-1}{n-1}}_{=2^{n-1}} \Big] \\\\ &\mathbf{=}& \mathbf{n\cdot 2^{n-1}} \\ \hline \end{array}\)
