Question

1+ 2.2+3.2²+4.2³+...100.2^99=?

ans- 99.2^100
​

Answer 1

$\large\underline{\sf{Solution-}}$

Given series is

$\rm :\longmapsto\:1 + 2.2 + 3. {2}^{2} + 4. {2}^{3} + - - - + 100. {2}^{99}$

Let assume that

$\rm :\longmapsto\:S = 1 + 2.2 + 3. {2}^{2} + 4. {2}^{3} + - - - + 100. {2}^{99}$

Its a series of sum of product of corresponding terms of two series, one AP and other GP

$\rm :\longmapsto\:1,2,3, - - - - - ,100$

and

$\rm :\longmapsto\:1,2, {2}^{2} , - - - - - , {2}^{99}$

So, multiply the given series by common ratio 2, we get

$\rm :\longmapsto\:2S = 1.2 + 2. {2}^{2} + 3. {2}^{3} + 4. {2}^{4} + - - - + 100. {2}^{100}$

On Subtracting these two equations, we get

$\rm :\longmapsto\: - S = 1 + 2 + {2}^{2} + - - + {2}^{99} - 100. {2}^{100}$

$\rm :\longmapsto\: - S = [1 + 2 + {2}^{2} + - - + {2}^{99}] - 100. {2}^{100}$

Now,

$\rm :\longmapsto\:1,2, {2}^{2} , - - - - - , {2}^{99} is \: a \: geometric \: progression$

whose

• First term, a = 1

• Common ratio, r = 2

• Number of terms, n = 100

We know, Sum of n terms of GP series having first term a and common ratio r respectively is

$\boxed{ \tt{ \: S_n \: = \: \frac{a \: ( {r}^{n} \: - \: 1)}{r \: - \: 1} \: }}$

So, using this result, we get

$\rm :\longmapsto\: - S = \dfrac{1( {2}^{100} - 1)}{2 - 1} - 100. {2}^{100}$

$\rm :\longmapsto\: - S = \dfrac{{2}^{100} - 1}{1} - 100. {2}^{100}$

$\rm :\longmapsto\: - S = {2}^{100} - 1- 100. {2}^{100}$

$\rm :\longmapsto\: - \: S \: = \: - \: 1 \: - \: 99. {2}^{100}$

$\rm :\longmapsto\: \: S \: = \: 1 \: + \: 99. {2}^{100}$

Hence,

$\boxed{ \tt{1 + 2.2 + 3. {2}^{2} + 4. {2}^{3} + - - + 100. {2}^{99} = 99. {2}^{100} + 1 \: }}$

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