For integers n and r, let (nr)={nCr,ifn≥r≥00,otherwise
The maximum value of k for which the sum k∑i=0(10i)(15k−i)+k+1∑i=0(12i)(13k+1−i) exists, is equal to
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Solution
BONUS QUESTION :
(1+x)10=10C0+10C1x+10C2x2+⋯+10C10x10 (1+x)15=15C0+15C1x+⋯+15Ck−1xk−1+15Ckxk+15Ck+1xk+1+⋯+15C15x15 k∑i=0(10Ci)(15Ck−i)=10C0⋅15Ck+10C1⋅15Ck−1+⋯+10Ck⋅15C0
Coefficient of xk in (1+x)25=25Ck
k+1∑i=0(12Ci)(13Ck+1−i)=12C0⋅13Ck+1+12C1⋅13Ck+⋯+12Ck+1⋅13C0
Coefficient of xk+1 in (1+x)25=25Ck+1
∴k∑i=0(10i)(15k−i)+k+1∑i=0(12i)(13k+1−i) =25Ck+25Ck+1 =26Ck+1
By the given definition of (nr),k can be as large as possible.