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Math Help - 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.
    (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.
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