Question

Find sum of series

$1 + 2(2) + 3( {2)}^{2} + 4( {2)}^{3} + ... + 100( {2}^{99} )$
​

Answer 1

Solution :

Let's assume that,

${ \longmapsto \sf S = 1+2(2)+3(2)^2+4(2)^3+...+100(2)^{99}\Big[Equation-1\Big]}$

Now multiplying this equation with 2 both sides.

${ \longmapsto \sf 2S = 2+2(2)^2+3(2)^3+...+100(2)^{100}\Big[Equation-2\Big]}$

Now subtracting equation 2 from equation 1

${ \longmapsto \sf S - 2S = 1+\Big[(2)+(2)^2+(2)^3+...+(2)^{99}\Big] - 100(2)^{100} }$

${ \longmapsto \sf S(1 - 2) = 1+\Big[(2)+(2)^2+(2)^3+...+(2)^{99}\Big] - 100(2)^{100} }$

Now, we can see that the terms inside brackets are forming a geometric sequence with first term 2, common ratio 2 and number of terms = 99.

We have a formula to find sum of n terms of a geometric sequence with first term a, common ratio r.

$\boxed{\mathfrak{ sum = \dfrac{a( {r}^{n} - 1) }{(r - 1)} }}$

By applying this formula, we get :

${ \longmapsto \sf - S= 1+2\left[ \dfrac{ {2}^{99} - 1 }{2 - 1} \right] - 100(2)^{100} }$

${ \longmapsto \sf - S= 1+{ {2}^{100} - 2 } - 100(2)^{100} }$

${ \longmapsto \sf - S= - 1+ {2}^{100} - 100(2)^{100} }$

${ \longmapsto \sf - S= - 1+ {2}^{100} ( 1 - 100)}$

${ \longmapsto \sf - S= - 1+ {2}^{100} ( - 99)}$

${ \longmapsto \sf - S= - 1 - {2}^{100} ( 99)}$

Now multiplying both sides with -1, we get :

$\underline {\boxed{ \sf S={2}^{100} ( 99) +1} }$

Therefore this is the required sum.

Answer 2

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

Given series is

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

Let Suppose that,

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

Now, the given series is product of corresponding terms of two series

$\red{\rm :\longmapsto\:1,2,3, - - - - ,100}$

and

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

So, Second series is an GP series with common ratio 2

So,

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

On multiply both sides by 2, we get

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

can be rewritten as

$\red{\rm :\longmapsto\:2S = [2 + 2. {2}^{2} + 3. {2}^{3} + 4. {2}^{3} + - - 99. {2}^{99}]+ 100. {2}^{100} - - (2)}$

On Subtracting equation (2) from equation (1), we get

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

We know that,

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

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

So,

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

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

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

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

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

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

$\bf\implies \:\boxed{ \tt{ \: S \: = \: 1 + 9 \times {2}^{100} \: \: }}$