If (1/(α+1)+1/(α+2)+… . .+1/(α+1012))-(1/(2 · 1)+1/(4 · 3)+1/(6 · 5)+… …+1/(2024 · 2023))=1/2024, then α is equal to ___________.

Solution

Maths sir · 4 steps

Split each 1/2k(2k-1) into two fractions, and the second bracket becomes a block of reciprocals.

1. Split each term. The general term of the second bracket is 1/2k(2k-1) for k = 1 to 1012. Partial fractions turn it into a difference of two reciprocals.

   1/2k(2k-1) = 1/(2k-1) - 1/2k

2. Rewrite the bracket. The second bracket becomes 1 - 1/2 + 1/3 - 1/4 + … - 1/2024. Add all the reciprocals up to 1/2024, then subtract twice the even ones. Twice the even ones is 1 + 1/2 + … + 1/1012, so what remains is 1/1013 + 1/1014 + … + 1/2024.
3. Find the first bracket. The first bracket must equal 1/1013 + … + 1/2024 + 1/2024. Since 1/2024 + 1/2024 = 1/1012, this is 1/1012 + 1/1013 + … + 1/2023. That is exactly 1012 terms, matching the first bracket.
4. So. The first term is 1/(α + 1) = 1/1012, so α = 1011. Check: 1/1012 - 1/2024 = 1/2024.

The idea to keep. An alternating sum 1 - 1/2 + … - 1/2n always equals 1/(n+1) + … + 1/2n. Spot it whenever you see terms like 1/2k(2k-1).

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