If 1 + (2 + ⁴⁹C₁ + ⁴⁹C₂ + ... + ⁴⁹C₄₉)(⁵⁰C₂ + ⁵⁰C₄ + ... + ⁵⁰C₅₀) is equal to 2ⁿ · m, where m is odd, then n+m is equal to __________.

Solution

Maths sir · 4 steps

Read the 2 as 1 + ⁴⁹C₀ and the first bracket becomes a full sum. Then it is a difference of squares.

1. First bracket. Split the 2 as 1 + ⁴⁹C₀. The bracket becomes 1 + (⁴⁹C₀ + ⁴⁹C₁ + … + ⁴⁹C₄₉) = 1 + 2⁴⁹.
2. Second bracket. The even-indexed coefficients of ⁵⁰Cᵣ add up to half of 2⁵⁰, which is 2⁴⁹. This bracket starts at ⁵⁰C₂, so take away ⁵⁰C₀ = 1. It equals 2⁴⁹ - 1.

   1 + (2⁴⁹+1)(2⁴⁹-1) = 1 + (2⁹⁸ - 1) = 2⁹⁸

3. Match the form. 2⁹⁸ = 2⁹⁸ · 1, and 1 is odd. So n = 98 and m = 1.
4. So. n + m = 98 + 1 = 99.

The common slip: 98. Reporting only n = 98 gives 98, not 99. The odd part m = 1 must also be added.

The idea to keep. When a binomial sum looks incomplete, add or remove the missing end terms to make a full sum. Remember that the even-indexed and odd-indexed ⁿCᵣ each add up to 2ⁿ⁻¹.

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