1.

Let Sn=1+q+q2+⋯+qn and Tn=1+(q+12)+(q+12)2+⋯+(q+12)n where q is a real number and q≠1. If 101C1+101C2⋅S1+⋯+ 101C101⋅S100=α T100, then α is equal to :

Answer»

Let Sn=1+q+q2++qn and Tn=1+(q+12)+(q+12)2++(q+12)n where q is a real number and q1. If 101C1+101C2S1++ 101C101S100=α T100, then α is equal to :



Discussion

No Comment Found