Question

Find the value of $x$ by evaluating the given series $1+\frac{1}{5}+\frac{1\times 3}{5\times 10}+\frac{1\times 3\times 5}{5\times 10\times 15}+...\infty =\sqrt{x},x\in Q$

• A. $\frac{2}{5}$
• B. $\frac{5}{3}$
• C. $\frac{3}{5}$
• D. $\frac{5}{2}$

Answer 1

Answer: (B)

$\frac{5}{3}$

Each term is of the following form:

$\frac{{\Pi }_{k=1}^n(2k-1)}{{\Pi }_{k=1}^n\left(5k\right)}$

$=\frac{{\Pi }_{k=1}^n(2k-1){\Pi }_{k=1}^n\left(2k\right)}{(n!)5^n{\Pi }_{k=1}^n\left(2k\right)}=\frac{\left(2n\right)!}{(n!)5^n(n!)2^n}=\frac{1}{{10}^n}{}^{2n}{C}_n$

$\therefore S=\sum _{n=0}^{\infty }\frac{1}{{10}^n}{}^{2n}{C}_n$

Now, since the result is the square root of a rational number, lets find $s^2$.

Using the Cauchy product (with $\frac{1}{10}$ as the independent variable), we get the following formula:

$S^2=\sum _{n=0}^{\infty }\frac{1}{{10}^n}\sum _{k=0}^n{}^{2n}{C}_k{}^{2(n-k)}{C}_{n-k}$

Now it can be shown that for all the whole numbers $n$, we have

$\sum _{k=0}^n{}^{2n}{C}_k{}^{2(n-k)}{C}_{n-k}=4^n$

Hence, we have

$S^2=\sum _{n=0}^{\infty }\frac{1}{{10}^n}{4}^n=\sum _{n=0}^{\infty }{\left(\frac{2}{5}\right)}^n$

$=\frac{1}{1-\frac{2}{5}}=\frac{5}{3}$

Hence, $x=\frac{5}{3}$