If $y=\frac25+\frac{1\cdot3}{2!}\left(\frac25\right)^2 +\frac{1\cdot3\cdot5}{3!}\left(\frac25\right)^3+\ldots$ then what is the value of $y^2+2y$?

Question

If $$y=\frac25+\frac{1\cdot3}{2!}\left(\frac25\right)^2 +\frac{1\cdot3\cdot5}{3!}\left(\frac25\right)^3+\ldots$$ then what is the value of $y^2+2y$?

This is a question from my coaching material in which binomial theorem, multinomial theorem and binomial theorem with fractional and negative indices are covered. How do I approach this problem? What is the pattern in it?

Answer 1

Recall the binomial series $(1+x)^\alpha=1+\alpha x+\frac{\alpha(\alpha-1)}{2!}+\dots$

Let $x=\frac{-4}{5}$ and $\alpha=\frac{-1}{2}$. Then $$ \sqrt{5} = \left(1-\frac{4}{5}\right)^{\frac{-1}{2}} = 1 +\frac{2}{5} + \frac{1\cdot3}{2!}\cdot\left(\frac{-1}{2}\right)^2\cdot\left(\frac{-4}{5}\right)^2+\dots =y+1 $$ so $y^2+2y=(y+1)^2-1=4$

How to arrive at this: in your question you stated that you want to use the binomial theorem. The thing stopping you is that the numerators are decreasing by two each time, rather than by one. This suggests that the exponent is a half-integer. Once you have this it's just a matter of fiddling to come up with the values of $x$ and $\alpha$.

Answer 2

Some hints:

We should look at the series $$f(x):=\sum_{k=1}^\infty{(2k-1)!!\over k!}\>x^k\ .$$ I suggest you bring ${1\over\sqrt{1-2x}}$ into the game. This should allow you to obtain a nice expression for $f(x)$.