Let S_n = 1+q+q^2++q^n and T_n = 1+q+12+(q+12)^2++(q+12)^n, where q is a real number, q1.

Question

Let $S_n = 1+q+q^2+\cdots+q^n$ and $T_n = 1+\dfrac{q+1}{2}+\left(\dfrac{q+1}{2}ight)^2+\cdots+\left(\dfrac{q+1}{2}ight)^n$, where $q$ is a real number, $q
e1$. If $\binom{101}{1}+\binom{101}{2}S_1+\binom{101}{3}S_2+\cdots+\binom{101}{101}S_{100} = \alpha T_{100}$, then $\alpha$ is equal to:

Accepted Answer

$S_n = \dfrac{1-q^{n+1}}{1-q}$, $T_{100} = \dfrac{1-\left(\frac{q+1}{2}\right)^{101}}{1-\frac{q+1}{2}} = \dfrac{2^{101}-(q+1)^{101}}{2^{100}(1-q)}$.

Using the binomial theorem to sum $\binom{101}{1}+\binom{101}{2}S_1+\cdots+\binom{101}{101}S_{100}$ and simplifying (via $\sum\binom{101}{k}q^{k-1}$-type identities) gives

$$\frac{1}{1-q}\left[2^{101}-(1+q)^{101}\right] = 2^{100}T_{100}.$$

So $\alpha = 2^{100}$.

Solutions · 1