Let 25r C_r and C_0+5C_1+9C_2++101C_25=2^25 k, then k is equal to: Answer: 51

Question

Let $\binom{25}{r}\equiv C_r$ and $C_0+5C_1+9C_2+\cdots+101C_{25}=2^{25}\cdot k$, then $k$ is equal to:

Answer: 51

Accepted Answer

The coefficient sequence $1,5,9,\ldots,101$ is $4r+1$ for $r=0,\ldots,25$.

$$\sum_{r=0}^{25}(4r+1)\binom{25}{r} = 4\sum r\binom{25}{r} + \sum\binom{25}{r} = 4(25\cdot2^{24}) + 2^{25} = 100\cdot2^{24}+2^{25}$$
$$= 2^{24}(100+2) = 2^{24}\times102 = 2^{25}\times51$$

$k = 51$

Solutions · 1