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Math Help - Number of integral terms in the expansion of (√6 + √10 + √15)^6 is?

  1. #1
    Super Member fardeen_gen's Avatar
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    Number of integral terms in the expansion of (√6 + √10 + √15)^6 is?

    Number of integral terms in the expansion of (√6 + √10 + √15)^6 is:

    A) 3
    B) 9
    C) 6
    D) 10
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  2. #2
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    Grandad's Avatar
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    Integral terms

    Hello fardeen_gen
    Quote Originally Posted by fardeen_gen View Post
    Number of integral terms in the expansion of (√6 + √10 + √15)^6 is:

    A) 3
    B) 9
    C) 6
    D) 10
    In the expansion of (a+b+c)^6, all the terms are in the form a^p b^q c^r, where p, q, r \ge 0 and p+q+r = 6.

    Now if a = \sqrt{6}, b = \sqrt {10} and c = \sqrt {15}, the only terms that have integer values are when p, q and r are even. (Note that abc = \sqrt{900} = 30, but the only way this combines with other even powers is in the term a^2b^2c^2; i.e. where p = q = r = 2.)

    So:

    • if p=0, q+r = 6, and there are 4 possible choices
    • if p = 2, q+r = 4, and there are 3 possible choices
    • if p = 4, q+r = 2: 2 choices
    • if p=6, q+r = 0: 1 choice

    Total number of terms, then, is 4+3+2+1 = 10. Answer: D.

    Grandad
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  3. #3
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    Hello, fardeen_gen!

    A variation of Grandad's explanation . . .


    Number of integral terms in the expansion of: \left(\sqrt6 + \sqrt{10} + \sqrt{15}\right)^6 is:

    . . A)\;3 \qquad B)\;9 \qquad C)\;6 \qquad D)\;10

    In the expansion of \left(\sqrt{6}+\sqrt{10}+\sqrt{15}\right)^6,

    . . the terms are of the form: {6\choose p,q,r}\left(\sqrt{6}\right)^p\left(\sqrt{10}\right  )^q\left(\sqrt{15}\right)^r . where p+q+r \:=\:6


    The term will be an integer if p,q,r are all even.

    There are three basic partitions . . .

    . . . \begin{array}{ccc}  \{p,q,r\} \:=\:\{0,0,6\} & \text{3 ways} \\ \{p,q,r\} \:=\:\{0,2,4\} & \text{6 ways} \\<br />
\{p,q,r\} \:=\:\{2,2,2\} & \text{1 way} \end{array}


    Therefore, there are 10 integral terms. (answer D)

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