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$\begingroup$

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 this

$(19).$ $$1+\frac{2n}3+\frac{2n\left(2n+2\right)}{3\cdot6}+\frac {2n\left(2n+2\right)\left(2n+4\right)}{3\cdot 6 \cdot 9}+\cdots \infty$$ $$=2^n \left(1+\frac{2n}3+\frac{2n\left(2n+2\right)}{3\cdot6}+\frac {2n\left(2n+2\right)\left(2n+4\right)}{3\cdot 6 \cdot 9}+\cdots \infty \right)$$

or give me some hints.

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    a) It's honestly too big of an image. b) The image shouldn't be the main portion of the question. c) please take the time to fully type out the question and at least post it right-side up. – 2017-01-29
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    Sorry ,,I laughed too much at what you said but I am a little knowledgable of the latex code so I am not aware how to type in latex.Whenever I try to type in latex it gives me goosebumps. – 2017-01-29
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    Lol, that used to happen to me too. If you haven't looked already, check here to see a full-blown $\LaTeX$ guide. – 2017-01-29
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    @Enlightened Why approve this faulty edit? – 2017-01-29
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    @No Thought-No Concept More thinking when editing would be a plus. – 2017-01-29

1 Answers 1

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The LaTeX version is wrong. I am completely referring to the photo originally taken.

We can rewrite the statement as $$\sum_{k=0}^{\infty} (\frac{2}{3})^k \binom{n+k-1}{k} = 2^n \sum_{k=0}^{\infty} (\frac{1}{3})^k \binom{n+k-1}{k}$$

We need to find a formula for $\sum_{k=0}^{\infty} a^k \binom{n+k-1}{k}$. Recall

$$\sum_{k=0}^{\infty} a^k = \frac{1}{1-a}$$

Differentiate both sides $n-1$ times with respect to $a$, we have

$$\sum_{k=0}^{\infty} a^k \frac{(n+k-1)!}{k!} = \frac{(n-1)!}{(1-a)^n} $$

We arrive at $$\sum_{k=0}^{\infty} a^k \binom{n+k-1}{k} = \frac{1}{(1-a)^n}\, , (|a|<1)$$

Plug in the corresponding values of $a$ in both sides, it is evident that

$$3^n = 2^n (\frac{3}{2})^n$$