Q.

The value of −15C1+2.15C2−3.15C3+......−15.15C15+14C1+14C3+14C5+......+14C11 is

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a

213−14

b

214

c

213−13

d

216−1

answer is A.

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Detailed Solution

S1=−15C1+2.15C2−......−15.15C15=∑r=115−1r.r.15Cr=∑r=115−1r.r.15rCr   14                                                                        =15-14C0+14C1−....−14C14=150=0                           Also    S2=14C1+14C3+......+14C11                                                                           = 14C1+14C3+....+14C11+14C13−14C13                 =213-14∴S1+S2=213−14
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