If _i=1^20(^20C_i-1^20C_i+^20C_i-1)^3 = k21, then k equals:

Question

If $\displaystyle\sum_{i=1}^{20}\left(\frac{^{20}C_{i-1}}{{}^{20}C_i+{}^{20}C_{i-1}}\right)^3 = \dfrac{k}{21}$, then $k$ equals:

Accepted Answer

$$\frac{{}^{20}C_{i-1}}{{}^{20}C_i+{}^{20}C_{i-1}} = \frac{{}^{20}C_{i-1}}{{}^{21}C_i} = \frac{i}{21}$$

$$\sum_{i=1}^{20}\left(\frac{i}{21}\right)^3 = \frac{1}{21^3}\sum_{i=1}^{20}i^3 = \frac{1}{21^3}\left(\frac{20\times21}{2}\right)^2 = \frac{100}{21}$$

$k=100$

Solutions · 1