I'm not going to expand this all the way!

Probability Level 3

$\large (1+x)(1+x^2)(1+x^3) \ldots (1+x^{99})(1+x^{100})$

What is the value of the coefficient of $x^9$ in the expansion above?

2 solutions

Murlidhar Sharma
Jun 3, 2015

This question was in 2015 JEE Advanced. You just have to find the number of ways the sum of whole numbers is equal to 9

i.e. $(0,9),(1.8),(2,7),(3,6),(4,5),(1,2,6),(1,3,5),(2,3,4)$

Manoj Kiran
Jun 1, 2015

The coefficient of $x^{9}$ will be same as the number of ways in which the sum of powers of x is 9.

Following are the number of ways in which the sum of powers of x is 9:

{(0,9), (1,8), (2,7), (3,6), (4,5)----->>5 ways.

{(1, 2, 6), (1, 3, 5), (2, 3, 4)}-------->>3 ways.

Therefore, the answer is 8.

Correct. Since you're already on the right track, let's make this problem harder!

Bonus question : With the same expression given, find the coefficient of $x^{21}$ .

Reply to the bonus question of Manoj Kiran

In the expansion of:

$\Pi_{k=1}^{100} (1 + x^k)$

the coefficient of $x^{21}$ equals 76.

The question is how to write 21 as the sum of distinct positive integers. All those possible distinct sums are in the figure below.

Therefore the coefficient of $x^{21} = 1 + 10 + 27 + 27 + 10 + 1 = 76$ .

Patrick Heebels - 6 years ago