The sum of 10 terms of the series
3/(1² × 2²) + 5/(2² × 3²) + 7/(3² × 4²) + .... is :

Solution

Maths sir · 4 steps

The numerator 2r+1 is exactly (r+1)² - r², so every term splits into a difference.

1. General term. The numerators are 3, 5, 7, …, so the rth numerator is 2r+1. The rth term is Tᵣ = (2r+1)/r²(r+1)².
2. Split the term. Notice that (r+1)² - r² = 2r+1. Put this in the numerator and break the fraction into two parts.

   Tᵣ = ((r+1)² - r²)/r²(r+1)² = 1/r² - 1/(r+1)²

3. Telescope. Adding from r = 1 to r = 10 gives (1/1² - 1/2²) + (1/2² - 1/3²) + … + (1/10² - 1/11²). Every middle term cancels, and only the first and the last are left.
4. So. S₁₀ = 1 - 1/11² = 1 - 1/121 = 120/121, which is option (b).

The idea to keep. When the numerator equals the difference of the two factors in the denominator, split the term into a difference. The sum of n terms is then the first piece minus the (n+1)th piece.

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