JEE-Advanced 2015 (23/40)

Probability Level 3

( 1 + x ) ( 1 + x 2 ) ( 1 + x 3 ) . . . ( 1 + x 100 ) \large (1+x)(1+x^2)(1+x^3)...(1+x^{100})

Find the coefficient of x 9 x^9 in the expansion above.


The answer is 8.

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Sandeep Bhardwaj by Sandeep Bhardwaj , 28, India

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1 solution

Vishnu Bhagyanath
Jun 21, 2015

An easy way to identify the answer is to find out the combinations through which we can arrive at x 9 x^9 .

Coefficient of x 9 x^9 would be in the following cases :

  1. x 9 x^ 9

  2. x 8 × x 1 x^8 \times x^1

  3. x 7 × x 2 x^7 \times x^2

  4. x 6 × x 3 x^6 \times x^3

  5. x 5 × x 4 x^5 \times x^4

  6. x 1 × x 2 × x 6 x^1 \times x^2 \times x^6

  7. x 1 × x 3 × x 5 x^1 \times x^3 \times x^5

  8. x 2 × x 3 × x 4 x^2 \times x^3 \times x^4

Since all the terms are of the forms ( 1 + x n ) (1+x^n) , our calculations become easy. To achieve a power of 9 using the above terms, all the remaining terms' 1 1 needs to be multiplied with each case. All other multiplications would lead to a power other than 9, which is currently not under observation.Doing so would give us a coefficient of 1 in each case. Since there are 8 cases, the coefficient would be 8.

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Is there a general solution?

Eric Kim - 4 years, 8 months ago

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