Question

Prove that, for all positive integers $n$, we have that \[ frac{2^{n+1} - 1}{n+1} = sum_{k=0}^n inom{n}{k} frac{1}{k+1}. \] ( extbf{Hint:} Use the Binomial Theorem on $(x + 1)^n$ and antiderivatives.)

Name: prove that for all positive integers n we have that frac2n1 1n1 sumk0n binomnk frac1k1 textbfhint use the binomial theorem on x 1n and antiderivatives
Uploaded: 2020-03-30T17:29:30-08:00
Duration: 1 min 52 s
Channel: Julian Wong
Description: prove that for all positive integers n we have that frac2n1 1n1 sumk0n binomnk frac1k1 textbfhint use the binomial theorem on x 1n and antiderivatives

          Prove that, for all positive integers $n$, we have that
		\[
			frac{2^{n+1} - 1}{n+1} = sum_{k=0}^n inom{n}{k} frac{1}{k+1}.
		\]
		(	extbf{Hint:} Use the Binomial Theorem on $(x + 1)^n$ and antiderivatives.)

Added by Vin A.

Higher Level Mathematics

Ibrahim Wazir, Tim Garry, Peter… 2012 Edition

Chapter 4

Instant Answer

Solved by Expert Julian Wong

03/30/2020

Step 1

Step 1:** Apply the Binomial Theorem to $(1+y)^n$: $$(1+y)^n = \sum_{k=0}^n \binom{n}{k} y^k$$ ** Show more…

Show all steps

Thanks for your feedback!

Prove that, for all positive integers $n$, we have that \[ frac{2^{n+1} - 1}{n+1} = sum_{k=0}^n inom{n}{k} frac{1}{k+1}. \] ( extbf{Hint:} Use the Binomial Theorem on $(x + 1)^n$ and antiderivatives.)