1•²⁵C₀ + 5•²⁵C₁ + 9•²⁵C₂ ..... 101•²⁵C₂₅ = 2²⁵ • k. Find the value of k.

Question

1•²⁵C₀ + 5•²⁵C₁ + 9•²⁵C₂ ..... 101•²⁵C₂₅ = 2²⁵ • k. Find the value of k. For the above question i figured out that it has a pattern i.e.

(4r+1)•²⁵Cᵣ

The solution to this question 51 but I dont know how to calculate it.

Answer 1

Let $\displaystyle S =1\cdot \binom{25}{0}+5\cdot \binom{25}{1}+9\cdot \binom{25}{2}+\cdots +101\cdot \binom{25}{25}\cdots (1)$

Writting in reverse order

$\displaystyle S =101\cdot \binom{25}{25}+97\cdot \binom{25}{24}+93\cdot \binom{25}{23}+\cdots +1\cdot \binom{25}{20}\cdots (2)$

Using $$\binom{n}{r}=\binom{n}{n-r}$$

$\displaystyle S =101\cdot \binom{25}{0}+97\cdot \binom{25}{1}+93\cdot \binom{25}{2}+\cdots +1\cdot \binom{25}{25}\cdots (3)$

Adding $(1)$ and $(3)$

$\displaystyle 2S =102\bigg[\binom{25}{0}+ \binom{25}{1}+\binom{25}{2}+\cdots\cdots +\binom{25}{25}\bigg]$

Finally $$2S=102\cdot 2^{25}\Longrightarrow S=51\cdot 2^{25}$$

Answer 2

Hint

$$(4r+1)\binom br=4 r\binom nr+\binom nr$$

$\displaystyle r\cdot\binom nr=\cdots=n\binom{n-1}{r-1}$

Now $(1+1)^m=\sum_{r=0}^m\binom mr$