Solve the recurrence relation

$\displaystyle \sum_{k=1}^n \binom{n}{k} a(k) = \frac{n}{n+1} $

I have two methods, one is general , by looking at first few terms and prove my deduction by induction .

I want to discuss the second one , I construct

$\displaystyle a(k) = c \int_0^1 [f(t)]^k~dt$ so we have

$\displaystyle c \int_0^1 \sum_{k=1}^n \binom{n}{k} [f(t)]^k~dt = \frac{n}{n+1} $

$\displaystyle c \int_0^1 ( [ 1 + f(t)]^n - 1) ~dt = \frac{n}{n+1}$

Let $\displaystyle c=-1 $ , then we obtain

$\displaystyle \int_0^1 [ 1 + f(t)]^n ~dt = \frac{1}{n+1} $

I know that $\displaystyle \int_0^1 t^n ~dt = \frac{1}{n+1} $

so i let $\displaystyle 1 + f(t) = t $ and i find that

$\displaystyle a(k) = -\int_0^1 (t-1)^k~dt = \frac{(-1)^{k+1}}{k+1}$

My problem is since $\displaystyle \int_0^1 [g(t)]^n~dt = \frac{1}{n+1} $ has many solutions , if we choose other solution , will we arrive at the same destination ?