Question

(10) 6. Determine whether the geometric series is convergent or divergent. If it converges, find its sum. 23n+2 10-3 no

Answer 1

we have

23η 1+2 -Σ 23:22 10*10-3 101-3 n=0 n=0

$\sum _{n=0}^{\infty } \frac{2^{3n+2}}{10^{n-3}}=\sum _{n=0}^{\infty } \frac{4(2^3)^n10^3}{10^{n}}$

$\sum _{n=0}^{\infty } \frac{2^{3n+2}}{10^{n-3}}=\sum _{n=0}^{\infty } \frac{4000(8)^n}{10^{n}}$

$\sum _{n=0}^{\infty } \frac{2^{3n+2}}{10^{n-3}}=\sum _{n=0}^{\infty } 4000\left (\frac{8}{10} \right )^{n}$

compare with geometric series,

$\sum_{n=0}^{\infty }a\left ( r \right )^{n}$

here a = 4000 and r = 8/10

hence |r| = 8/10 < 1

therefore, the series is converges by the geometric series test.

the formula of the sum of geometric series is,

$\sum_{n=0}^{\infty }a\left ( r \right )^{n}=\frac{a}{1-r}$

we can say that,

$\sum _{n=0}^{\infty } \frac{2^{3n+2}}{10^{n-3}}=\frac{4000}{1-\left (\frac{8}{10} \right )}$

$\sum _{n=0}^{\infty } \frac{2^{3n+2}}{10^{n-3}}=\frac{4000}{\frac{10-8}{10}}$

$\sum _{n=0}^{\infty } \frac{2^{3n+2}}{10^{n-3}}=\frac{4000}{\frac{2}{10}}$

$\sum _{n=0}^{\infty } \frac{2^{3n+2}}{10^{n-3}}=\frac{4000}{\frac{1}{5}}$

$\sum _{n=0}^{\infty } \frac{2^{3n+2}}{10^{n-3}}=4000(5)$

$\sum _{n=0}^{\infty } \frac{2^{3n+2}}{10^{n-3}}=20000$