Question

Prove that the sum of the series $1+3+7+15+31+....+$ upto $100$ terms is $2^{101}-102$.

Answer 1

$1+3+7+15+\mathrm{31...}$

$=(2-1)+(4-2)+(8-1)+(16-1)+(32-1)...$

$=(2+4+8+16+\mathrm{32...})-(1+1+1+1+...)$

Sum of $2+4+8+16+\mathrm{32...}45$ to $100$ term

$=$ sum of C.P $2,4,8,\mathrm{16...}$

$=\frac{a\left(r^{n-1}\right)}{r-1}=\frac{2\left(2^{100-1}\right)}{2-1}$

$=2^{101}-2$

Sum of $1+1+1+1$ up to 100 terms $=1\times 100=100$

$\therefore$ Total sum of series $=2^{101}-2-100$

$=2^{101}-102$