If 1/(2 . 3¹⁰) + 1/(2² . 3⁹) + ..... + 1/(2¹⁰ . 3) = K/(2¹⁰ . 3¹⁰), then the remainder when K is divided by 6 is :

Solution

Maths sir · 4 steps

Clear the denominator first, and K turns into the familiar sum that equals 3¹⁰ - 2¹⁰.

1. Clear the denominator. The r-th term is 1/(2ʳ · 3¹¹⁻ʳ) for r = 1 to 10. Multiply both sides by 2¹⁰ · 3¹⁰, and each term becomes 2¹⁰⁻ʳ · 3ʳ⁻¹.

   K = 2⁹ + 2⁸ · 3 + 2⁷ · 3² + … + 3⁹ = Σₖ₌₀⁹ 3ᵏ 2⁹⁻ᵏ

2. Telescope it. Each term can be written as a difference: 3ᵏ⁺¹ 2⁹⁻ᵏ - 3ᵏ 2¹⁰⁻ᵏ = 3ᵏ 2⁹⁻ᵏ(3 - 2) = 3ᵏ 2⁹⁻ᵏ. Adding from k = 0 to 9, the middle terms cancel and only 3¹⁰ - 2¹⁰ is left. This is the identity aⁿ - bⁿ = (a - b)(aⁿ⁻¹ + aⁿ⁻²b + … + bⁿ⁻¹) with a - b = 1.
3. Find the remainder. K = 59049 - 1024 = 58025. Since 6 × 9670 = 58020, the remainder is 58025 - 58020 = 5.
4. So. The remainder when K is divided by 6 is 5, option (d).

The idea to keep. When the powers of two bases slide in opposite directions term by term, the sum is (aⁿ - bⁿ)/(a - b). Spot that pattern before you start adding.

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