Question

$1+2\cdot 2+3\cdot 2^2+4\cdot 2^3+.....+100\cdot 2^{99}$

• A. $S={50502}^{100}$
• B. $S={5050.2}^{99}$
• C. $S={5050.2}^{101}$
• D. $S={5050.2}^{98}$

Answer 1

Answer: (B)

$S={5050.2}^{99}$

Consider the given series.

$1+2.2+{3.2}^2+{4.2}^3+{5.2}^4+.......+{100.2}^{99}$

The series mentioned above is a $A.G.P$ series i.e arithmetic and geometric series.

The $n^{th}$ term for this series will be

$=n{.2}^{n-1}$

Now,

$=\frac{n(n+1)}{2}.\frac{{1.2}^{n-1}}{2-1}$

Since, $n=100$

Therefore,

$=\frac{100(100+1)}{2}.\frac{{1.2}^{100-1}}{2-1}$

$=\frac{100\left(101\right)}{2}.{1.2}^{99}$

$=50\left(101\right){.2}^{99}$

$={5050.2}^{99}$

Hence, this is the answer.