-1
$\begingroup$

It looks like a easy question ,but when you try to take $2$ common you are stuck,I don't know what to do in this question...How to take out $2^n$ from the LHS ,I am confused... Please help me prove that $ $$1+\frac{2n}3+\frac{2n\left(2n+2\right)}{3\cdot6}+\frac {2n\left(2n+2\right)\left(2n+4\right)}{3\cdot 6 \cdot 9}+\cdots \infty$$ $$=2^n \left(1+\frac{n}3+\frac{n\left(n+2\right)}{3\cdot6}+\frac {n\left(n+2\right)\left(n+4\right)}{3\cdot 6 \cdot 9}+\cdots \infty \right)$

or give me some hints.

  • 1
    The RHS is meaningless – 2017-01-29
  • 0
    @user254665 In fact, the RHS is meaningful (see the answer by Robert Israel) but not equal to the LHS... – 2017-01-29
  • 0
    @Robert Israel Should the RHS be $=2^n \left(1+\frac{n}3+\frac{n\left(n+1\right)}{3\cdot6}+\frac {n\left(n+1\right)\left(n+2\right)}{3\cdot 6 \cdot 9}+\cdots \right)$, it would be equal to the LHS. – 2017-01-29

1 Answers 1

3

It's not true.

$$ 1 + \frac{2n}{3} + \frac{(2n)(2n+2)}{3\cdot 6} + \ldots = 3^n $$ and thus (substituting $n/2$ for $n$) $$ 1 + \frac{n}{3} + \frac{n(n+2)}{3 \cdot 6} + \ldots = 3^{n/2}$$

The first formula, BTW, can be obtained from the binomial series for $ (1-2 z/3)^{-n}$ at $z=1$.

  • 0
    [+1} It took me a little while to see... the evidence.... – 2017-01-29