Consider the next succession and prove by induction

Question

The exercise says:

Knowing the next succession,

$$a_1=1$$ $$a_{n+1}=\frac{2a_n}{(2n+2)(2n+1)}, n>=1$$

Prove by induction that $a_n=\frac{2^n}{(2n)!}$

What I've done so far is prove for $n=1$

For $n=1$:

$$a_1=\frac{2^1}{(2\times1)!}=\frac{2}{2}=1$$ Which is correct.

Therefore, to prove the induction:

$$a_{n+1}=\frac{2\times a_{n+1}}{(2(n+1)+2)(2n+1)}=$$ $$=\frac{2}{(2(n+1)+2)(2n+1)}\frac{2^{n+1}}{(2(n+1))!}=$$ $$=\frac{2^{n+2}}{(2n+4)(2n+3)}\times\frac{1}{(2(n+1))!}=$$ $$=\frac{2^{n+2}}{4n^2+6n+18n+12}=$$ $$=\frac{2^{n+2}}{4n^2+24n+12}=$$ $$=\frac{2\times2^{n+1}}{2(2n^2+12n+6)}=$$ $$=\frac{2^{n+1}}{2n^2+12n+6}$$

Is this correct? Do I have to simplify even more?

Thanks

Answer 1

No, it is not correct. What you need to show is that, assuming $$a_n=\frac{2^n}{(2n)!},$$ then $$a_{n+1} = \frac{2a_n}{(2n+2)(2n+1)} = \frac{2^{n+1}}{\big(2(n+1)\big)!}.$$ So $$\frac{2a_n}{(2n+2)(2n+1)} = \frac{2}{(2n+2)(2n+1)}\frac{2^n}{(2n)!}=\dots$$ Try to follow from there.

Answer 2

When you wrote that$$a_{n+1}=\frac{2\times a_{n+1}}{(2(n+1)+2)(2n+1)},$$I suppose that you meant to write $a_{n+2}$ as the LHS.

Anyway, this is inconclusive. If you were assuming that $a_{n+1}=\frac{2^{n+1}}{(2(n+1))!}$, your conclusion should have been that $a_{n+2}=\frac{2^{n+2}}{(2(n+2))!}$.

Note that$$\frac{a_{n+1}}{a_n}=\frac2{(2n+2)(2n+1)}\tag1$$and that$$\frac{\frac{2^{n+1}}{(2(n+1))!}}{\frac{2^n}{(2n)!}}=\frac{2\times(2(n+1))!}{(2n)!}=\frac2{(2n+2)(2n+1)}=(1).$$Can you take it from here?