Prove by induction this sequence

Question

Prove by induction, that

$$a_n=\frac{n+1}{n-1}(a_1+a_2+\ldots+a_{n-1})$$

is $$a_n=(n+1)2^{n-2}\;,$$ where $a_1=1$.

I have tried the indution step, but cannot succeed.

Answer 1

Hint: your induction step is equivalent to prove that

$$\sum_{k=1}^{n-1}(k+1)2^k = (n-1)2^n $$

Now the sums $\sum_{k=1}^{n-1}2^k$ and $\sum_{k=1}^{n-1}k2^k$ are easy to find.

Answer 2

Hint

Note that

$$\begin{align} a_{n+1}&\\ &=\frac{n+2}{n}(a_1+\cdots a_{n-1}+a_n)\\ &=\frac{n+2}{n}(a_1+\cdots a_{n-1})+\frac{n+2}{n}a_n\\ &=\frac{n+2}{n}\frac{n-1}{n+1}\frac{n+1}{n-1}(a_1+\cdots a_{n-1})+\frac{n+2}{n}a_n\\ &=\frac{n+2}{n}\frac{n-1}{n+1}a_n+\frac{n+2}{n}a_n\\ &=\frac{n+2}{n}\left(\frac{n-1}{n+1}+1\right)a_n\\ &=\frac{n+2}{n}\frac{2n}{n+1}a_n\\ &=\frac{2(n+2)}{n+1}a_n.\end{align}$$

Answer 3

If $a_n=(n+1)2^{n-2}$, then

$$a_1+a_2+\cdots+a_{n-1}={n-1\over n+1}a_n=(n-1)2^{n-2}$$

Using both expressions as the inductive hypothesis, we get

$$\begin{align} a_{n+1}&={n+2\over n}(a_1+a_2+\cdots+a_{n-1}+a_n)\\ &={n+2\over n}\left((n-1)2^{n-2}+(n+1)2^{n-2} \right)\\ &=(n+2)2^{n-1}\\ &=((n+1)+1)2^{(n+1)-2} \end{align}$$