A probability problem by Atul Kumar Ashish

$\begin{aligned} S & = 8 {40 \choose 0} + 15 {40 \choose 1} + 22 {40 \choose 2} + \cdots + 288 {40 \choose 40} \\ & = \sum_{n=0}^{40} (7n+8) {40 \choose n} \\ & = 7 \sum_{n=0}^{40} {40 \choose n}n + 8 \sum_{n=0}^{40} {40 \choose n} \\ & = 7 \frac d{dx} (1+{\color{#3D99F6}x}) ^{40} + 8 (1+{\color{#3D99F6}x})^{40} & \small \color{#3D99F6} \text{where }x = 1 \text{ (see Note.)} \\ & = 280 (1+{\color{#3D99F6}x}) ^{39} + 8 (1+{\color{#3D99F6}x})^{40} & \small \color{#3D99F6} \text{Putting back }x = 1 \\ & = 35\cdot 2^{42} + 2\cdot 2^{42} \\ & = 37\cdot 2^{42} \end{aligned}$

$\implies a + b = 37 + 42 = \boxed{79}$

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Note: We note that:

$\begin{aligned} \sum_{k=0}^n {n \choose k}x^k & = (1+x)^n & \small \color{#3D99F6} \text{Putting }x = 1 \\ \implies \sum_{k=0}^n {n \choose k} & = 2^n \\ \frac d{dx} \sum_{k=0}^n {n \choose k}x^k & = \frac d{dx} (1+x)^n & \\ \sum_{k=0}^n k{n \choose k}x^{k-1} & = n(1+x)^{n-1} \\ \sum_{k=1}^n k{n \choose k}x^{k-1} & = n(1+x)^{n-1} & \small \color{#3D99F6} \text{Putting }x = 1 \\ \implies \sum_{k=1} ^n k{n \choose k} & = 2^{n-1}n \end{aligned}$

$8{40 \choose 0}+15{40 \choose 1}+22{40 \choose 2}+29{40 \choose 3}+\cdots+288{40 \choose 40} = S$ $288{40 \choose 0}+281{40 \choose 1}+274{40 \choose 2}+267{40 \choose 3}+\cdots+8{40 \choose 40} = S$ $296{40 \choose 0}+296{40 \choose 1}+296{40 \choose 2}+296{40 \choose 3}+\cdots+296{40 \choose 40} = 2S$ $2S = 296\left({40 \choose 0}+{40 \choose 1}+{40 \choose 2}+{40 \choose 3}+\cdots+{40 \choose 40}\right).$ $2S = 296(1+1)^{40}$ $S = 148\cdot 2^{40}$ $S = 37\cdot 2^{42}$ $a+b = 37+42 = \boxed{79}$