Question

prove that the sum to n terms of the series 11+103+1005+.......... is 10 / 9 (10to the power n - 1)+n (sq).

Answer 1

Hey

The given Series is :

$(10 + 1) + ( {10}^{2} + 3) + ...$

Which can be reduced as :

$\sum( {10}^{n + 1} + 2n + 1) - 1$

Breaking it :

$= \sum {10}^{n + 1} + \sum (2n + 1) - 1$
$= \frac{1}{9} ( {10}^{n + 2} - 1) + {n}^{2} - 1$

Answer 2

Step-by-step explanation:

see the attachment

hope this will help you

thank you