<s>[INST] Express $1 + \frac {1}{2} \binom{n}{1} + \frac {1}{3} \binom{n}{2} + \dotsb + \frac{1}{n + 1}\binom{n}{n}$ in a simplifed form I need to express $$1 + \frac {1}{2} \binom{n}{1} + \frac {1}{3} \binom{n}{2} + \dotsb  + \frac{1}{n + 1}\binom{n}{n}$$ in a simplified form.
So I used the identity $$(1+x)^n=1 + \binom{n}{1}x + \binom{n}{2}x^2 + \dotsb   + \binom{n}{n}x^n$$
Now on integrating both sides and putting $x=1$.
I am getting $$\frac{2^{n+1}}{n+1}$$ is equal to the given expression.But the answer in my book is  $$\frac{2^{n+1}-1}{n+1}.$$
Where does that -1 term in the numerator come from?
 [/INST] Remember when you integrate both sides, there will be the constant of integration. You need to first find this constant of integration. That's where the $-1$ comes from. In actual fact, it is $\frac{-1}{n+1}$ but looks like $-1$ because it's been absorbed into the numerator.
You may like to try $x=0$ to find this $+C$ value...
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