\(\text{Simplifying the expression we will get }\)
\(\displaystyle \frac{4}{20}+\frac{4.7}{10^2(3!)}+\frac{4.7.10}{10^3(4!)}+\cdots\)
\(\text{Multiplying and dividing by 10 we will get}\)
\(\displaystyle 10\left(\frac{\frac{1}{3}\left(\frac{1}{3}+1\right)3^2}{10^2 .2!}+\frac{\frac{1}{3}\left(\frac{1}{3}+1\right) \left(\frac{1}{3}+2\right).3^3}{10^3.3!} \cdots\right)\)
\(\displaystyle 10\left(\frac{\frac{1}{3}\left(\frac{1}{3}+1\right) \frac{3^2}{10^2}}{2!}+\frac{\frac{1}{3}\left(\frac{1}{3}+1\right) \left(\frac{1}{3}+2\right).\frac{3^3}{10^3}}{3!} \cdots\right) \cdots \cdots\;\; (1)\)
\(\text{We have a result which we will use here to solve this expression}\)
\(\displaystyle \left(1-x\right)^{-n}=1+nx+\frac{n(n+1)x^2}{2!}+\frac{n(n+1)(n+2)}{3!}+\cdots \cdots (2)\)
\(\text{Above equation can be written as}\)
\(\displaystyle \left(1-x\right)^{-n}-nx-1=+\frac{n(n+1)x^2}{2!}+\frac{n(n+1)(n+2)}{3!}+\cdots\;\; (3)\)
\(\text{So using (2) in (1) we can say that n = -1/3 and x=3/10 }\)
\(\text{Using above we can write (1) with the help of (3) As}\)
\(\displaystyle 10 \left[\left(1-\frac{3}{10}\right)^{\frac{-1}{3}}-\frac{1}{3}\times \frac{3}{10}-1 \right] \cdots \cdots \;\;\;\;(4)\)
\(\text{Further (4) can be simplified and we will get}\)
\(\displaystyle 10 \left[\left(\frac{7}{10}\right)^{\frac{-1}{3}}-\frac{1}{10}-1 \right]\)
\(\text{ANSWER :} \;\;\;\; 10\left(\frac{10}{7}\right)^{\frac{1}{3}} -11\)
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