Evaluate $\sum\limits_{k = 1}^{11} {\left( {2 + {3^k}} \right)} $
$\sum\limits_{k = 1}^{11} {\left( {2 + {3^k}} \right) = } \sum\limits_{k = 1}^{11} {\left( 2 \right) + } \sum\limits_{k = 1}^{11} {\left( {{3^k}} \right) = 22 + } \sum\limits_{k = 1}^{11} {{3^k}} $ .........$(1)$
$\sum\limits_{k = 1}^{11} {{3^k} = {3^1} + {3^2} + {3^3} + ........ + {3^{11}}} $
The terms of this sequence $3,3^{2}, 3^{3} \ldots \ldots$ forms a $G.P.$
$S_{n}=\frac{a\left(r^{n}-1\right)}{r-1}$
$\Rightarrow S_{n}=\frac{3\left[(3)^{11}-1\right]}{3-1}$
$\Rightarrow S_{n}=\frac{3}{2}\left(3^{11}-1\right)$
$\therefore \sum\limits_{k = 1}^{11} {{3^k}} = \frac{3}{2}\left( {{3^{11}} - 1} \right)$
Substituting this value in equation $(1),$ we obtain
$\sum\limits_{k = 1}^{11} {\left( {2 + {3^k}} \right) = 22 + \frac{3}{2}\left( {{3^{11}} - 1} \right)} $
If $2^{10}+2^{9} \cdot 3^{1}+28 \cdot 3^{2}+\ldots+2 \cdot 3^{9}+3^{10}=S -211$ then $S$ is equal to
The product $(32)(32)^{1/6}(32)^{1/36} ...... to\,\, \infty $ is
Let $P(x)=1+x+x^2+x^3+x^4+x^5$. What is the remainder when $P\left(x^{12}\right)$ is divided by $P(x)$ ?
A $G.P.$ consists of an even number of terms. If the sum of all the terms is $5$ times the sum of the terms occupying odd places, then the common ratio will be equal to
If the sum of an infinite $G.P.$ and the sum of square of its terms is $3$, then the common ratio of the first series is