Question

The sum of the series $i+2+3i+4\cdots$ upto $100$ terms where $i=\sqrt{-1}$ equal p + iq.
Then number of positive integral divisors of p-q equals

• A. 4
• B. 6
• C. 10
• D. 20

Answer 1

The correct option is B 6

i+2+3i+4...100 terms

$i(1+3+\mathrm{5...95})+(2+4+6+...100)$

$i[{\sum }_{100}^1\frac{n(n+1)}{2}-\sum$ even terms]+2[1+2+...50]

$i[\frac{100\times 101}{2}-\sum$ even terms]+$2x\frac{n+1}{2}{|}_{n=50}$

$\Rightarrow$ $[50\times 101-\sum$ even terms]+$50\times 51$

$\Rightarrow i[50\times 101-50\times 51]+50\times 51$

$\Rightarrow 2500i+2550$

$\therefore$ p = 2550 & q = 2500

$\therefore$ p-q=50.divisors : 1, 2, 5, 10, 25, 50

$\therefore$ No .of divisors $\Rightarrow 6.$