JEE Main 2019 — Sequences And Series Question with Solution

Let S = $$\sum _{k=1}^{20}k\frac{1}{2^k}$$

$$\Rightarrow$$ S = $$\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+.......+\frac{20}{2^{20}}$$ .......(1)

This is an Arithmetic Geometric Sequence. Here numerator is in A.P and denominator is in G.P.

To solve Arithmetic Geometric Sequence, we multiply the series S with the common ratio of G.P. Here common ratio of G.P is $$\frac{1}{2}$$.

By multiplying (1) with $$\frac{1}{2}$$ we get,

$$\frac{S}{2}=\frac{1}{2^2}+\frac{2}{2^3}+\frac{3}{2^4}+\frac{4}{2^5}+.......+\frac{19}{2^{20}}+\frac{20}{2^{21}}$$ ....(2)

Subtract (2) from (1),

S - $$\frac{S}{2}$$ = $$\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+.......+\frac{1}{2^{20}}-\frac{20}{2^{21}}$$

$$\Rightarrow$$ $$\frac{S}{2}$$ = $$\frac{\frac{1}{2}\left(1-\frac{1}{2^{20}}\right)}{1-\frac{1}{2}}$$ - $$\frac{20}{2^{21}}$$

$$\Rightarrow$$ S = $$2\left(1-\frac{1}{2^{20}}\right)-\frac{20}{2^{20}}$$

$$\Rightarrow$$ S = $$2-\frac{2}{2^{20}}-\frac{20}{2^{20}}$$

$$\Rightarrow$$ S = $$2-\frac{22}{2^{20}}$$

$$\Rightarrow$$ S = $$2-\frac{11}{2^{19}}$$