Question

Sum $1+\frac{2}{2}+\frac{3}{2^2}+\frac{4}{2^3}+....$ to $n$ terms.

Answer 1

Let $S=1+\frac{2}{2}+\frac{3}{2^2}+\frac{4}{2^3}+\frac{5}{2^4}.....\frac{n}{2^{n-1}}$
Multiplying both sides by $\frac{1}{2}$, we get
$S\frac{1}{2}=0+\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}...........\frac{n-1}{2^{n-1}}+\frac{n}{2^n}$
Subtracting both eqns, we get
$S\frac{1}{2}=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}...........\frac{1}{2^{n-1}}-\frac{n}{2^n}$
Clearly above series is a G.P upto second last term with $r=\frac{1}{2}$
Then, $S\frac{1}{2}=\frac{1[{\left(\frac{1}{2}\right)}^n-1]}{\frac{1}{2}-1}-\frac{n}{2^n}$
$\Rightarrow S=4[1-{\left(\frac{1}{2}\right)}^n]-\frac{n}{2^{n-1}}$