Using the concept of the binomial expansion solve: $\sum_{k=0}^{n} \frac{1}{k+1} \binom{n}{k}$

Question

I really have trouble starting these. Usually once I've begun I'm ok, but any help starting would be most appreciated.

$$ \sum_{k=0}^{n} \frac{1}{k+1} \binom{n}{k} $$

Answer 1

Hint: $$ \frac{1}{k+1}\binom{n}{k} = \frac{n!}{(n-k)!(k+1)!} = \frac{n!}{[(n+1)-(k+1)]!(k+1)!} $$ Try to express this using binomial coefficients.

Answer 2

Hint: consider the derivative of

$$\sum_{k=0} ^n \frac 1{k+1}\binom nk X^{k+1} $$ details:

$$f(X) = \sum_{k=0} ^n \frac 1{k+1}\binom nk X^{k+1}\\ f'(X) = \sum_{k=0} ^n \binom nk X^{k} = (1+X)^n\implies f(X) = \frac1{n+1}(1+X)^{n+1} \\ \sum_{k=0} ^n \frac 1{k+1}\binom nk = f(1)-f(0) = \frac{2^{n+1}-1}{n+1} $$

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Another way: $$\frac 1{k+1}\binom nk X^{k+1} = \frac {n!}{(n-k)! (k+1)!} = \frac1{n+1}\frac {(n+1)!}{(n-k)! (k+1)!} = \frac1{n+1} \binom {n+1}{k+1} $$

Then, $$\sum_{k=0} ^n \frac 1{k+1}\binom nk =\frac1{n+1}\sum_{k=0} ^n \binom{n+1}{k+1} = \frac{2^{n+1}-1}{n+1} $$

Answer 3

Hint

$(1+x)^n=\displaystyle\binom{n}{0}+\binom{n}{1}x+\binom{n}{2}x^2+....+\binom{n}{n}x^n$

and

$\displaystyle\int{y^n}=\displaystyle\frac{1}{n+1}y^{n+1}+C$

$\displaystyle\frac{(1+x)^{n+1}}{n+1}+C=\displaystyle\binom{n}{0}x+\frac{1}{2}\binom{n}{1}x^2+\frac{1}{3}\binom{n}{2}x^3+....+\frac{1}{n+1}\binom{n}{n}x^{n+1}$