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- Thread starter MarkFL
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- Mar 10, 2012

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Evaluate the given sum:

\(\displaystyle S_n=\sum_{k=0}^n\left(\frac{1}{k+1}{n \choose k} \right)\)

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While this is an very straightforward method (much more elegant than my pre-calculus approach), your end result is incorrect, but only because you did not apply the FTOC correctly.

- Mar 10, 2012

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Ah.. I see. I should get $\frac{2}{n+1} - \frac{1}{n+1} = \frac{1}{n+1}$. Where have I committed a mistake? I cannot find it.While this is an very straightforward method (much more elegant than my pre-calculus approach), your end result is incorrect, but only because you did not apply the FTOC correctly.

And what was your solution. I'd be interested in a pre-calculus solution.

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You should get:Ah.. I see. I should get $\frac{2}{n+1} - \frac{1}{n+1} = \frac{1}{n+1}$. Where have I committed a mistake? I cannot find it.

And what was your solution. I'd be interested in a pre-calculus solution.

I will post my solution soon, but I want to give others a chance to post their solutions first.

- Mar 10, 2012

- 835

haha. I wasn't doing the algebra right. How silly of me.You should get:

I will post my solution soon, but I want to give others a chance to post their solutions first.

- Mar 31, 2013

- 1,346

So sum (nCk)/(k+1) = 1/(n+1) (sum (n+1 C k) - (n+1 C 0))

= 1/(n+1) (sum (n+1Ck) - 1)

= 1/(n+1) ( 2^(n+1) – 1)

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\(\displaystyle 2^{n+1}=(1+1)^{n+1}=\sum_{k=0}^{n+1}{n+1 \choose k}\)

Next, apply the identities:

\(\displaystyle {n \choose 0}=1\)

\(\displaystyle {n+1 \choose r}=\frac{n+1}{r}{n \choose r-1}\)

and so we have:

\(\displaystyle 2^{n+1}=1+(n+1)\sum_{k=1}^{n+1}\left(\frac{1}{k}{n \choose k-1} \right)\)

Subtracting through by $1$ and re-indexing the sum, we have:

\(\displaystyle 2^{n+1}-1=(n+1)\sum_{k=0}^{n}\left(\frac{1}{k+1}{n \choose k} \right)\)

Dividing through by $n+1$ and using \(\displaystyle S_n=\sum_{k=0}^n\left(\frac{1}{k+1}{n \choose k} \right)\) we find:

\(\displaystyle S_n=\frac{2^{n+1}-1}{n+1}\)