Question

Find the sum to n terms of the series
11 + 102 + 1003 + 10004 +...:

• A. $({10}^n-1)+\frac{n(n+1)}{2}$
• B. $\frac{10}{9}({10}^n-1)+\frac{n(n+1)}{2}$
• C. ${10}^n+n^2-1$
• D. None of these

Answer 1

The correct option is B $\frac{10}{9}({10}^n-1)+\frac{n(n+1)}{2}$

Best way to solve is to go through options.
Put n = 1 in options, you should get sum = 11.
For n = 2, you should get sum = 113.
Only option (b) is true for both, hence is the correct option.

Alternatively:
= 11 + 102 + 1003 + ...
= (10 + 1) + (100 + 2) + (1000+ 3)
= (10 + 100 + 1000 +...) + (1 + 2 + 3 +...)