Question

Find the sum of series $1.2+2.2^2+3.2^3+...+100.2^{100}$.

Answer 1

Given series is $s=1.2+{2.2}^2+{3.2}^3....+{100.2}^{100}$ where s is sum of series

multiplying the whole series with 2 we get the new series as

$2s={1.2}^2+{2.2}^3+{3.3}^4+.....100{.2}^{101}$

now subtracting these two weg get

$2s-s=s=\left[\right(1-2){.2}^2+(2-3){.2}^3+(3-4){.2}^4+....(99-100\left){.2}^{100}\right]+{100.2}^{101}-1.2$

$s=[2^2+2^3+2^4...{.2}^{100}]+2.({100.2}^{100}-1)$

first is sum of G.P with r=2,a=2^2=4,n=99

therefore the first part is $\frac{a(r^n-1)}{r-1}=\frac{4(2^{99}-1)}{2-1}=2.2^{100}-4$

so the sum is ${102.2}^{101}-6$