If (1^15C_0+1^15C_1)(1^15C_1+1^15C_2) (1^15C_12+1^15C_13)= ^13^14C_0^14C_1 ^14C_12, then 3…

Question

If $(\frac{1}{{}^{15}{C}_{0}}+\frac{1}{{}^{15}{C}_{1}})(\frac{1}{{}^{15}{C}_{1}}+\frac{1}{{}^{15}{C}_{2}})\ldots (\frac{1}{{}^{15}{C}_{12}}+\frac{1}{{}^{15}{C}_{13}})=\frac{{\alpha }^{13}}{{}^{14}{C}_{0}{}^{14}{C}_{1}\cdots {}^{14}{C}_{12}}$, then $30\alpha $ is equal to
$____$ .

Answer: 32

Accepted Answer

For each factor, use
$$\frac{1}{{}^{n}C_r}+\frac{1}{{}^{n}C_{r+1}} = \frac{{}^{n}C_{r+1}+{}^{n}C_r}{{}^{n}C_r\cdot {}^{n}C_{r+1}} = \frac{{}^{n+1}C_{r+1}}{{}^{n}C_r\cdot {}^{n}C_{r+1}}$$

Since ${}^{n+1}C_{r+1} = \dfrac{n+1}{r+1}\,{}^nC_r$, this simplifies to

$$\frac{1}{{}^{n}C_r}+\frac{1}{{}^{n}C_{r+1}} = \frac{n+1}{n}\cdot\frac{1}{{}^{n-1}C_r}\cdot\frac{1}{r+1}\cdot(r+1)=\frac{n+1}{n}\cdot\frac{1}{{}^{n+1}C_{r+1}}\cdot{}^{n+1}C_{r+1}$$

More directly, applying this with $n=15$ for each of the 13 factors gives each factor equal to $\dfrac{16}{15}\cdot\dfrac{1}{{}^{14}C_r}$. Multiplying all 13 factors:

$$\left(\frac{1}{{}^{15}C_0}+\frac{1}{{}^{15}C_1}\right)\cdots\left(\frac{1}{{}^{15}C_{12}}+\frac{1}{{}^{15}C_{13}}\right) = \left(\frac{16}{15}\right)^{13}\cdot\frac{1}{{}^{14}C_0\cdot{}^{14}C_1\cdots{}^{14}C_{12}}$$

Comparing with the given form, $\alpha = \dfrac{16}{15}$, so

$$30\alpha = 30\times\frac{16}{15} = 32$$

Solutions · 1