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Sequences and Series question

2015 · Shift 2 · Q28
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Sequences and Series question

2015 · Shift 2 · Q28

JEE AdvancedMathematicsSequences and SeriesNumerical+4 / −1
The coefficient of x9{x^9}x9 in the expansion of (1 + x) (1 +x2){x^2)}x2)(1 +x3{x^3}x3) .... (1+x100)(1 + {x^{100}})(1+x100) is
Numerical answer
View written solutionFree

Correct answer: 8

  1. We need the coefficient of x9x^9x9 in
(1+x)(1+x2)(1+x3)⋯(1+x100).(1+x)(1+x^2)(1+x^3) \cdots (1+x^{100}).(1+x)(1+x2)(1+x3)⋯(1+x100).
  1. In the product ∏k=1100(1+xk)\prod_{k=1}^{100}(1+x^k)∏k=1100​(1+xk), choosing either 111 or xkx^kxk from each factor means:
  • every term corresponds to selecting some distinct numbers from 1,2,3,…,1001,2,3,\dots,1001,2,3,…,100,
  • the exponent of xxx is the sum of the selected numbers.

So, the coefficient of x9x^9x9 equals the number of ways to write 999 as a sum of distinct positive integers.

Since any part bigger than 999 cannot be used, we only need partitions of 999 into distinct parts.

  1. List all partitions of 999 into distinct positive integers:
999 8+18+18+1 7+27+27+2 6+36+36+3 6+2+16+2+16+2+1 5+45+45+4 5+3+15+3+15+3+1 4+3+24+3+24+3+2

These are all possible distinct-part partitions of 999.

  1. Therefore, the coefficient of x9x^9x9 is
8.8.8.
  1. Comparing with the stored correct answer: stored answer = 888, which matches our result.
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