The greatest integer less than or equal to the sum of first 100 terms of the sequence 1/3,5/9,19/27,65/81, ...... is equal to ___________.

Solution

Maths sir · 4 steps

Write each term as 1 minus a power of 2/3, and the sum splits into 100 minus a small GP.

1. Spot the pattern. Each numerator is 1 less than the denominator minus a power of 2: 1 = 3 - 2, 5 = 9 - 4, 19 = 27 - 8, 65 = 81 - 16. So the nth term is (3ⁿ - 2ⁿ)/3ⁿ = 1 - (2/3)ⁿ.
2. Split the sum. Adding 100 such terms gives 100 ones, minus a GP with first term 2/3 and common ratio 2/3.

   S₁₀₀ = 100 - (2/3(1 - (2/3)¹⁰⁰))/(1 - 2/3) = 98 + 2(2/3)¹⁰⁰

3. Size the leftover. 2(2/3)¹⁰⁰ is positive but extremely small, far less than 1. So S₁₀₀ lies just above 98 and below 99.
4. So. The greatest integer less than or equal to S₁₀₀ is 98.

The common slip: 97. Starting the GP at (2/3)⁰ = 1 makes its sum close to 3 instead of 2, giving 97.

The idea to keep. When numerators look odd, compare each one with its denominator. A term like 1 - rⁿ turns the series into a count of ones minus a GP, and the tiny r¹⁰⁰ part only decides the floor.

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