The value of Σᵣ₌₀²⁰ ⁵⁰⁻ʳC₆ is equal to:

Solution

Maths sir · 4 steps

Turn each ⁿC₆ into a difference of two ⁿC₇ terms, and the whole sum telescopes.

1. List the terms. As r goes from 0 to 20, 50 - r runs from 50 down to 30. So the sum is ³⁰C₆ + ³¹C₆ + … + ⁵⁰C₆, which is 21 terms.
2. Pascal's rule. From ⁿC₆ + ⁿC₇ = ⁿ⁺¹C₇, each term can be rewritten as a difference.

   ⁿC₆ = ⁿ⁺¹C₇ - ⁿC₇

3. Telescope. Add this for n = 30, 31, …, 50. Each middle term ³¹C₇, ³²C₇, …, ⁵⁰C₇ appears once with a plus sign and once with a minus sign, so they cancel. Only ⁵¹C₇ from the last term and -³⁰C₇ from the first term remain.
4. So. The sum equals ⁵¹C₇ - ³⁰C₇, which is option (b).

The idea to keep. A sum of ⁿCₖ with n running and k fixed always telescopes through ⁿCₖ = ⁿ⁺¹Cₖ₊₁ - ⁿCₖ₊₁: the top goes up by one, the bottom goes up by one.

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