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Q.

In the expansion of 1+x+7x11, the term not containing x is

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a

∑r=05 11Cr11−rC2r7r

b

∑r=15 11Cr11−rC11−2r7r

c

∑r=05 11Cr11−rC11−2r7r

d

∑r=15r c 11

answer is C.

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Detailed Solution

1+x+7x11=7+x+x2x11So, we have to find the coefficient of x11 in 7+x+x211Now 7+x+x211 =(7+x)+x211=11C0(7+x)11+11C1(7+x)10x2+11C2(7+x)9x4…+… ∴ Required coefficient =11C0+i1C1×10C9×7+11C2×9C7×72+11C3×8C5×73+11C4×7C3×74+11C5×6C1×75 =∑r=05 11Cr11−rC11−2r7r

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