) ∵‌‌(1+x)10=‌10C0+‌10C1x+‌10C2x2+...+‌10C10x10 and (1+x)15=‌15C0+‌15C1x+‌15C2x2+...+‌15C15x15 ∴
k
∑
i=0
(‌10Ci)(‌15Ck−i)=‌10C0⋅‌15Ck+‌10C1⋅‌15Ck−1+...+‌10Ck‌15C0 ⇒ Coefficient of xk in (1+x)25=‌25Ck Also,
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 ⇒‌‌‌25Ck+‌25Ck+1=‌26Ck+1 So, ‌26Ck+1 always exists. Now k can be as larger as possible.