Question-9600

Question Number 9600 by Sopheak last updated on 20/Dec/16

Commented by prakash jain last updated on 20/Dec/16

k^{t h} Term = \underset{i = 0}{\sum^{k - 1}} 18 \times (10^{2 i}) = \frac{18 (10^{2 k} - 1)}{99}

S_{n} = \frac{2}{11} \underset{k = 1}{\sum^{n}} (10^{2 k} - 1)

= \frac{2}{11} (\frac{10^{2} (10^{2 n} - 1)}{99} - n)