Q49 / 160 · AP EAMCET 2006 · Mathematics
$1+\frac{2}{4}+\frac{2 \cdot 5}{4 \cdot 8}+\frac{2 \cdot 5 \cdot 8}{4 \cdot 8 \cdot 12}+\frac{2 \cdot 5 \cdot 8 \cdot 11}{4 \cdot 8 \cdot 12 \cdot 16}+\ldots \ldots$ is equal to :
- A$4^{-2 / 3}$
- Correct answerB$\sqrt[3]{16}$
- C$\sqrt[3]{4}$
- D$4^{3 / 2}$
Solution
Let $S=1+\frac{2}{4}+\frac{2 \cdot 5}{4 \cdot 8}+\frac{2 \cdot 5 \cdot 8}{4 \cdot 8 \cdot 12}+\ldots$
On comparing with
$(1+x)^n=1+n x+\frac{n(n-1)}{2 !} x^2+\ldots$
we get
$n x=\frac{2}{4}$ $\ldots$ (i)
and $\quad \frac{n(n-1)}{2 !} x^2=\frac{2 \cdot 5}{4 \cdot 8}$ $\ldots$ (ii)
From (i) and (ii)
$\frac{\frac{n(n-1)}{2 !} x^2}{n^2 x^2}=\frac{\frac{2 \cdot 5}{4 \cdot 8}}{\frac{2}{4} \cdot \frac{2}{4}}$
$\Rightarrow \quad \frac{n-1}{n}=\frac{5}{2}$
$\Rightarrow \quad 2 n-2=5 n$
$\Rightarrow \quad n=-\frac{2}{3}$
On Putting the value of $n$ in Eq. (i), we get
$-\frac{2}{3} x=\frac{2}{4}$
$\Rightarrow \quad x=-\frac{3}{4}$
$\therefore S=(1+x)^n=\left(1-\frac{3}{4}\right)^{-2 / 3}=\left(\frac{1}{4}\right)^{-2 / 3}=\sqrt[3]{16}$
Hence, the correct answer is option: 'B'.