# Thread: Arithmetic Progressions and Binomial Theorem?

1. ## Arithmetic Progressions and Binomial Theorem?

Let {(1 + x^2)^2}.(1 + x)^4 = ∑(r = 0 to 8)ar.x^r, then show that a1, a2, a3 are in A.P.

NOTE: In ar, r is a subscript.

2. Originally Posted by fardeen_gen
Let {(1 + x^2)^2}.(1 + x)^4 = ∑(r = 0 to 8)ar.x^r, then show that a1, a2, a3 are in A.P.

NOTE: In ar, r is a subscript.
$(1+x^2)(1+x^2)(1+x)(1+x)(1+x)(1+x)$

The $a_1$ is the number of ways of getting $x$ coefficient. If you look at the above expanded product the only way to get $x$ coefficient is when from the first two binomals we choose $1$ and from one of the last four binomials we choose $x$. There are 4 ways of doing that so $a_1=4$.

The $a_2$ is the number of ways of getting $x^2$ coefficint. If you look at the above expanded product you can get $x^2$ if you choose $x^2$ in first binomial and let others be $1$ or if you choose first one to be $1$ second binomal to be $x^2$ and all others to be $1$. This already gives us two ways. But another way is to choose $1$ from the first two binomials and then choose two $x$'s from the last four. Thus, there are $2+{4\choose 2} = 8$ ways of doing that so $a_2 = 8$.

The $a_3$ term can be got by choosing $x^2$ from first binomial and pairing it with one of the $x$'s from the last four binomials and so there are $4$ ways already. Similar argument involving $x^2$ in the second binomial gives us another $4$ ways. And finally we can choose $1$ from the first two binomials and then 3 $x$'s from the last four binomials. Therefore, $a_3 = 4 + 4 + {4\choose 3} = 12$.