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For any m $\ge$ 1. Prove that $$\sum_{j=0}^m\frac{1}{j+1}{m \choose j} = \frac{1}{m+1}(2^{m+1} -1 )$$

Attempt: I tried using subsets after rearranging the equation. Then even though the R.H.S gives the no of substes of n+1 excluding empty. I am unable to get a interpretation for L.H.S

Amrita
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2 Answers2

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HINT:

$$\frac1{j+1}\binom{m}j=\frac1{m+1}\binom{m+1}{j+1}$$

(This is an identity worth knowing; it comes up quite often.)

Brian M. Scott
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Start with $$ (1+x)^m=\sum_{j=0}^{m}{m\choose j}x^j $$ and integrate both sides from $0$ to $1$.

carmichael561
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  • Sorry should I integrate with respect to m. The RHS ${m \choose j}$ is constant. How do you do for $x^j$ . My calculus is weak – Amrita Sep 10 '16 at 20:23
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    No, integrate with respect to $x$. $\int_0^1x^j;dx=\frac{1}{j+1}$, and $\int_0^1(1+x)^m;dx=\frac{2^{m+1}-1}{m+1}$. – carmichael561 Sep 10 '16 at 20:24