Question:
If $\mathrm{C}_{\mathrm{r}} \equiv{ }^{25} \mathrm{C}_{\mathrm{r}}$ and $\mathrm{C}_{0}+5 \cdot \mathrm{C}_{1}+9 \cdot \mathrm{C}_{2}+\ldots+(101) \cdot \mathrm{C}_{25}=2^{25} \cdot k$
then $k$ is equal to________.
Solution:
$\sum_{r=0}^{25}(4 r+1){ }^{25} C_{r}=4 \sum_{r=0}^{25} r \cdot{ }^{25} C_{r}+\sum_{r=0}^{25}{ }^{25} C_{r}$
$=4 \sum_{r=1}^{25} r \times \frac{25}{r}{ }^{24} C_{r-1}+2^{25}=100 \sum_{r=1}^{25}{ }^{24} C_{r-1}+2^{25}$
$=100.2^{24}+2^{25}=2^{25}(50+1)=51.2^{25}$
Hence, by comparison $k=51$
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