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Thread: Arithmetic Progressions and Binomial Theorem?

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    Super Member fardeen_gen's Avatar
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    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.
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    Quote Originally Posted by fardeen_gen View Post
    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.
    $\displaystyle (1+x^2)(1+x^2)(1+x)(1+x)(1+x)(1+x)$

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

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

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