I'm Not Going To Expand Everything!

Algebra Level 4

Find the coefficient of x 8 x^8 in the expansion of

( x − 1 ) ( x − 2 ) ( x − 3 ) ⋯ ( x − 9 ) ( x − 10 ) . (x-1)(x-2)(x-3) \cdots (x-9)(x-10) .


The answer is 1320.

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Anshu Garg by anshu garg , 18, India

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1 solution

Relevant wiki: Vieta's Formula - Forming Quadratics

The product can be expressed into a polynomial as ∏ n = 1 10 ( x − n ) = x 10 − a 9 x 9 + a 8 x 8 − . . . + a 0 \displaystyle \prod_{n=1}^{10} (x-n) = x^{10} - a_9x^9 + a_8x^8 - ... + a_0 . It has roots x i = i x_i = i . Using Vieta's formula, the coefficient of x 8 x^8 , a 8 a_8 is given by:

a 8 = ∑ i ≠ j 10 x i x j = 1 2 ( ∑ i 10 ∑ j 10 x i x j − ∑ k = 1 10 x k 2 ) = 1 2 ( ( ∑ n = 1 10 n ) 2 − ∑ n = 1 10 n 2 ) = 1 2 ( ( 10 ( 10 + 1 ) 2 ) 2 − 10 ( 10 + 1 ) ( 2 ⋅ 10 + 1 ) 6 ) = 5 5 2 − 385 2 = 1320 \begin{aligned} a_8 & = \sum_{i \ne j}^{10} x_ix_j \\ & = \frac 12 \left( \sum_{i}^{10} \sum_{j}^{10} x_i x_j - \sum_{k=1}^{10} x_k ^2 \right) \\ & = \frac 12 \left(\left(\sum_{n=1}^{10} n \right)^2 - \sum_{n=1}^{10} n^2 \right) \\ & = \frac 12 \left(\left( \frac {10(10+1)}2 \right)^2 - \frac {10(10+1)(2\cdot 10+1)}6 \right) \\ & = \frac {55^2 - 385}2 = \boxed{1320} \end{aligned}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

FYI Avoid having your notation do double duty. You had the roots denoted as a i a_i initially, which I've edited into x i x_i .

Calvin Lin Staff - 4 years, 7 months ago

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