Binomial Coefficient Problem
- Thread starter currybeef
- Start date
\(\displaystyle (1+x)^n = \sum \limits_{k=0}^n \dbinom{n}{k}x^k \\~\\
\left . \sum \limits_{k=0}^n \dbinom{n}{k} = (1+x)^n \right|_{x=1} = 2^n\\~\\
\text{differentiate}\\~\\
n(1+x)^{n-1} = \sum \limits_{k=1}^n k \dbinom{n}{k}x^{k-1} \\~\\
\left . \sum \limits_{k=1}^n k \dbinom{n}{k} = n(1+x)^{n-1} \right|_{x=1} = n2^{n-1} \\~\\
\text{The sum given in part (a) is}\\~\\
\sum \limits_{k=2}^n (k-1)\dbinom{n}{k} = \\~\\
\sum \limits_{k=2}^n k\dbinom{n}{k} - \sum \limits_{k=2}^n \dbinom{n}{k} = \\~\\
(n2^{n-1}-n) - (2^n - n - 1) = \\~\\
n2^{n-1}-2^n + 1 = \\~\\
(n-2)2^{n-1}+1
\)
\left . \sum \limits_{k=0}^n \dbinom{n}{k} = (1+x)^n \right|_{x=1} = 2^n\\~\\
\text{differentiate}\\~\\
n(1+x)^{n-1} = \sum \limits_{k=1}^n k \dbinom{n}{k}x^{k-1} \\~\\
\left . \sum \limits_{k=1}^n k \dbinom{n}{k} = n(1+x)^{n-1} \right|_{x=1} = n2^{n-1} \\~\\
\text{The sum given in part (a) is}\\~\\
\sum \limits_{k=2}^n (k-1)\dbinom{n}{k} = \\~\\
\sum \limits_{k=2}^n k\dbinom{n}{k} - \sum \limits_{k=2}^n \dbinom{n}{k} = \\~\\
(n2^{n-1}-n) - (2^n - n - 1) = \\~\\
n2^{n-1}-2^n + 1 = \\~\\
(n-2)2^{n-1}+1
\)
Steven G
Elite Member
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- Dec 30, 2014
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On your 2nd line, the derivative is wrong. Why is that x in the numerator factored out?
Oh, thanks for pointing out.