The term independent of x in the expansion of160−x881⋅2×2−3×26 is equal to

The term independent of x in the expansion of

160x8812x23x26 is equal to

  1. A

    -108

  2. B

    -72

  3. C

    -36

  4. D

    36

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    Solution:

    Given expansion is 160x8812x23x26

    =1602x23x26x8812x23x26

    So its general term is

    Tr+1=1606Cr2x26r3x2rx8816Cr2x26r3x2r=1606Cr(2)6r(3)rx122r2r1816Cr(2)6r(3)rx122r2r+8=1606Cr(2)6r(3)rx124r1816Cr(2)6r(3)rx204r

    For this term to be independent of x,put r=3 in 

    1st part and r = 5 in 2nd part.

    From ,  (i), Tr+1=160×20×8(27)+181×6×2×243

    =72+36=36

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