The value of −15C1+2⋅15C2−3⋅15C3+⋯−15⋅15C15+14C1+14C3+14C5+⋯+14C11 is
Solutions · 1
Official solution
Let S1=r=1∑15(−1)rr⋅15Cr=15r=1∑15(−1)r14Cr−1=15(−14C0+14C1−⋯−14C14)=15×0=0
Let S2=14C1+14C3+⋯+14C11=(14C1+14C3+⋯+14C13)−14C13=213−14S1+S2=0+213−14=213−14