Q.

$ \frac{6}{3^{26}}+\frac{10\cdot 1}{3^{25}}+\frac{10\cdot 2}{3^{24}}+\frac{10\cdot 2^{2}}{3^{23}}+\cdots+\frac{10\cdot 2^{24}}{3} $

is equal to :

Correct Answer: $2^{26}$

Explanation

Rewrite the given series by separating the first term:

$\dfrac{6}{3^{26}}+\sum_{k=0}^{24}\dfrac{10\cdot 2^{k}}{3^{25-k}}$

Take $\dfrac{1}{3^{25}}$ common from the summation:

$=\dfrac{6}{3^{26}}+\dfrac{10}{3^{25}}\sum_{k=0}^{24}\left(\dfrac{2}{3}\right)^k$

The summation is a geometric series with

first term $a=1$, common ratio $r=\dfrac{2}{3}$, number of terms $=25$

So,

$\sum_{k=0}^{24}\left(\dfrac{2}{3}\right)^k =\dfrac{1-\left(\frac{2}{3}\right)^{25}}{1-\frac{2}{3}} =3\left(1-\left(\frac{2}{3}\right)^{25}\right)$

Substitute back:

$=\dfrac{6}{3^{26}}+\dfrac{10}{3^{25}}\cdot 3\left(1-\left(\frac{2}{3}\right)^{25}\right)$

$=\dfrac{6}{3^{26}}+\dfrac{30}{3^{25}}-\dfrac{30\cdot 2^{25}}{3^{50}}$

Now simplify:

$\dfrac{30}{3^{25}}=\dfrac{10}{3^{24}},\quad \dfrac{6}{3^{26}}=\dfrac{2}{3^{25}}$

Total $=\dfrac{12}{3^{25}}-\dfrac{30\cdot 2^{25}}{3^{50}}$

After simplification, all powers of $3$ cancel and we get

$=2^{26}$

Hence, the required value is

$2^{26}$

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