+0  
 
0
65
4
avatar

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.

Guest Oct 31, 2018
 #1
avatar
+1

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.

Guest Oct 31, 2018
 #2
avatar+20116 
+3

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.

 

\(\begin{array}{|rcll|} \hline && \mathbf{1\dbinom n1 + 2\dbinom n2 + 3\dbinom n3 + 4\dbinom n4 + \ldots + n\dbinom nn} \\\\ &=& 1\cdot\dfrac{n!}{(n-1)!\cdot 1} + 2\cdot\dfrac{n!}{(n-2)!\cdot 1\cdot 2} + 3\cdot\dfrac{n!}{(n-3)!\cdot 1\cdot 2\cdot 3} \\ &&+ 4\cdot\dfrac{n!}{(n-4)!\cdot 1\cdot 2\cdot 3\cdot 4} + \ldots + n\cdot\dfrac{n!}{(n-n)!\cdot 1\cdot 2\cdot \ldots \cdot(n-1)\cdot n } \\\\ &=& \dfrac{n!}{(n-1)!\cdot 0!} + \dfrac{n!}{(n-2)!\cdot 1!} + \dfrac{n!}{(n-3)!\cdot 2!} + \dfrac{n!}{(n-4)!\cdot 3!} \\ && +\ldots + \dfrac{n!}{(n-n)!\cdot(n-1)! } \\\\ &&\boxed{n!=n\cdot(n-1)!} \\\\ &=& \dfrac{n\cdot(n-1)!}{(n-1)!\cdot 0!} + \dfrac{n\cdot(n-1)!}{(n-2)!\cdot 1!} + \dfrac{n\cdot(n-1)!}{(n-3)!\cdot 2!} + \dfrac{n\cdot(n-1)!}{(n-4)!\cdot 3!} \\ && +\ldots + \dfrac{n\cdot(n-1)!}{(n-n)!\cdot(n-1)! } \\\\ &=& n\cdot \Big[ \dfrac{ (n-1)!}{(n-1)!\cdot 0!} + \dfrac{ (n-1)!}{(n-2)!\cdot 1!} + \dfrac{ (n-1)!}{(n-3)!\cdot 2!} + \dfrac{ (n-1)!}{(n-4)!\cdot 3!} \\ && +\ldots + \dfrac{ (n-1)!}{(n-n)!\cdot(n-1)! } \Big] \\\\ &=& n\cdot \Big[ \underbrace{\dbinom{n-1}{0}+\dbinom{n-1}{1}+\dbinom{n-1}{2}+\dbinom{n-1}{3} +\ldots + \dbinom{n-1}{n-1}}_{=2^{n-1}} \Big] \\\\ &\mathbf{=}& \mathbf{n\cdot 2^{n-1}} \\ \hline \end{array}\)

 

 

laugh

heureka  Nov 1, 2018
edited by heureka  Nov 1, 2018
 #3
avatar
+1

\(x^2=64=x=8\)

Guest Nov 1, 2018
 #4
avatar
+1

\(\sqrt{900}*2=60, \sqrt{900}=30\)

pls solve

Guest Nov 1, 2018

29 Online Users

avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.