6/3¹² + 10/3¹¹ + 20/3¹⁰ + 40/3⁹ + ... + 10240/3 = 2ⁿ . m, where m is odd, then m . n is equal to ____________.

Solution

Maths sir · 4 steps

Add the terms one at a time from the left and watch each running sum collapse to a neat power of 2.

1. Spot the GP. From the second term on, each term is (10 × 2ᵏ)/3¹¹⁻ᵏ for k = 0 to 10, since 10240 = 10 × 2¹⁰. The first term 6/3¹² = 2/3¹¹ does not fit this pattern, so keep it separate.
2. Start adding. 2/3¹¹ + 10/3¹¹ = 12/3¹¹ = 4/3¹⁰. Then 4/3¹⁰ + 20/3¹⁰ = 24/3¹⁰ = 8/3⁹. Each new term has the same denominator as the running sum, and the total keeps shrinking to a power of 2.

   2ᵏ⁺¹/3¹¹⁻ᵏ + (10 × 2ᵏ)/3¹¹⁻ᵏ = (12 × 2ᵏ)/3¹¹⁻ᵏ = 2ᵏ⁺²/3¹⁰⁻ᵏ

3. Last term. Just before the last term (k = 10) the running sum is 2¹¹/3 = 2048/3. Adding 10240/3 gives 12288/3 = 4096 = 2¹².
4. So. 2¹² = 2ⁿ × m with m odd gives n = 12 and m = 1, so m × n = 12.

The idea to keep. When a series has one odd-looking first term, add it to the next term and check the running sum. It often telescopes into a clean pattern. Or sum the GP part and see what the extra term cancels.

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