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Q.

The value of −15C1+2⋅15C2−3⋅15C3+….15⋅15C15+14C1+14C3+14C5+…+14C11 is

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a

214

b

213-13

c

216-1

d

213-14

answer is D.

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Detailed Solution

S1=15C1+2⋅15C2−……−1515C15=∑r=115 (−1)r⋅r15C2=15∑r=115 (−1)r14Cr−1=15−14C0+14C1−……−14C14=15(0)=0S2=14C114C3+…..+14C11= 14C1+14C3+…….14C11+14C13−14C13=213−14=S1+S2=213−14
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