Question

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Answer 1

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Answer 2

Answer:

Step-by-step explanation:

The given series is an Arithmetico-Geometric series i.e. it is a series obtained by multiplying terms of an AP and a GP.

Consider LHS of the given series and let it be 'S'.

\sf S = 1 \cdot 2 + 2\cdot2^2 + 3\cdot2^3 +... + (n-1)\cdot2^{n-1} + n\cdot2^nS=1⋅2+2⋅22+3⋅23+...+(n−1)⋅2n−1+n⋅2n

Let it be equation 1

Multiply both sides with 2

\sf 2S = 1 \cdot 2^2 + 2\cdot2^3 + 3\cdot2^4 +... + (n-1)\cdot2^{n} + n\cdot2^{n+1}2S=1⋅22+2⋅23+3⋅24+...+(n−1)⋅2n+n⋅2n+1

Let it be equation 2

Now subtract equation 1 from equation 2

\small{\sf-S = 1\cdot2 +\bigg[ (2\cdot2^2 - 1\cdot 2^2) + (3\cdot2^3 - 2\cdot2^3) +...+ (n\cdot2^n - \{n-1\}\cdot2^{n})\bigg]- n\cdot2^{n+1}}−S=1⋅2+[(2⋅22−1⋅22)+(3⋅23−2⋅23)+...+(n⋅2n−{n−1}⋅2n)]−n⋅2n+1

\small{\sf-S = 1\cdot2 +\bigg[ 1\cdot2^2 +1\cdot2^3 +...+ 1\cdot2^n \bigg]- n\cdot2^{n+1}}−S=1⋅2+[1⋅22+1⋅23+...+1⋅2n]−n⋅2n+1

\small{\sf-S = \underbrace{\sf(2 + 2^2 + 2^3 + ...+ 2^n)}_{GP\, with\, a\,=\,r\,=\,2}-n\cdot2^{n+1}}−S=GPwitha=r=2(2+22+23+...+2n)−n⋅2n+1

Sum of n terms of a GP is given by:

\boxed{\tt S_n = \dfrac{a(r^n-1)}{r-1}}Sn=r−1a(rn−1)

\small{\sf-S =\dfrac{2(2^n-1)}{2-1}-n\cdot2^{n+1}}−S=2−12(2n−1)−n⋅2n+1

$$\small{\sf-S ={2(2^n-1)-n\cdot2^{n+1}}$$

\small{\sf-S =2^{n+1}-2-n\cdot2^{n+1}}−S=2n+1−2−n⋅2n+1

\small{\sf-S =2^{n+1}(-n+1)-2}−S=2n+1(−n+1)−2

\small{\sf-S =-(n-1)2^{n+1}-2}−S=−(n−1)2n+1−2

Multiply both sides with -ve 1

\small{\sf S =(n-1)2^{n+1}+2}S=(n−1)2n+1+2

Hence the required result is proved that:

\boxed{\tt 1 \cdot 2 + 2\cdot2^2 + 3\cdot2^3 +... + (n-1)\cdot2^{n-1} + n\cdot2^n = (n-1)2^{n+1} + 2}1⋅2+2⋅22+3⋅23+...+(n−1)⋅2n−1+n⋅2n=(n−1)2n+1+2

Step-by-step explanation:

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