Let S = 2 + 6/7 + 12/7² + 20/7³ + 30/7⁴ + ...... Then 4S is equal to

Solution

Maths sir · 5 steps

Numerators 2, 6, 12, 20 have second differences constant, so multiply by 1/7 and subtract twice.

1. Spot the pattern. The numerators are 1·2, 2·3, 3·4, 4·5, …, so the nth term is n(n+1)/7ⁿ⁻¹. Their differences 4, 6, 8, ... form an AP, so one shift-and-subtract will not be enough. You need two.
2. First subtraction. Write S/7 = 2/7 + 6/7² + 12/7³ + … and subtract it from S. This gives 6/7S = 2 + 4/7 + 6/7² + 8/7³ + …. Call the right side T.
3. Second subtraction. Do the same to T: T - T/7 = 2 + 2/7 + 2/7² + …, which is an infinite GP with first term 2 and ratio 1/7.

   6/7T = 2/(1-1/7) = 7/3 ⇒ T = 49/18

4. Back to S. 6/7S = T = 49/18, so S = 49/18 × 7/6 = 343/108.
5. So. 4S = 343/27 = (7/3)³, which is option (c).

The idea to keep. Each multiply-by-r-and-subtract lowers the degree of the numerator by one. A quadratic numerator needs two rounds before you reach a plain GP.

Answer from the NTA final answer key. Solution by PrepWiser.