Combinations

The coefficient of $x^7$ of each factor $a_k$ is as below:

$\overbrace{(1+x)^7}^{a_0 = \binom 77} + \overbrace{(1+x)^8}^{a_1=\binom 87} + \overbrace{(1+x)^9}^{a_2=\binom 97} + \overbrace{(1+x)^{10}}^{a_3=\binom {10}7} + \cdots + \overbrace{(1+x)^N}^{a_{N-7}=\binom N7}$

Therefore, the coefficient of $x^7$ for the expression is $\displaystyle A_{N-7} = \sum_{k=0}^{N-7} a_k = \sum_{k=0}^{N-7} \binom {k+7}7 = \sum_{k=0}^{N-7} \binom {k+7}k$ . From the first few $A_n$ , it appears that $A_n = \dbinom {n+8}n$ . Let us prove the claim to be true for all $n \ge 0$ by induction.

Proof: For $n=0$ , $A_0 = a_0 = \dbinom 70 = 1 = \dbinom 80$ indicating that the claim is true for $n=0$ . Now assuming the claim is true for $n$ , then:

$\begin{aligned} A_{n+1} & = A_n + \binom {n+8}{n+1} \\ & = \color{#3D99F6} \binom {n+8}n + \binom {n+8}{n+1} & \small \color{#3D99F6} \text{By Pascal's formula} \\ & = \color{#3D99F6} \binom {n+9}{n+1} \end{aligned}$

Therefore, the claim is also true for $n+1$ and hence true for all $n \ge 0$ . Then we have $A_{10^7-7} = \dbinom {10^7+1}{10^7-7} = \boxed{\dbinom {10000001}8}$ .

$(1+x)^{n}$ = $nCr$ × $x^{r}$ ,

so, $x^{7}$ in $(1+x)^{n}$ = $nC7$ × $x^{7}$

so

the coefficient of $x^{8}$ in $(1+x)^{8}$ + $(1+x)^{9}$ + $(1+x)^{10}$ +...+ $(1+x)^{10000000}$

= $7C7$ + $8C7$ + $9C7$ + $10C7$ +...+ $10000000C7$

= $8C8$ + $8C7$ + $9C7$ +...+ $10000000C7$ ------------------------------------ ( because $nCn$ = $rCr$ = $kCk$ = 1 )

= $9C8$ + $9C7$ + $10C7$ +...+ $10000000C7$ ------------------------------------ ( because $nCn$ = $rCr$ = $kCk$ = 1 )

= $10C8$ + $10C7$ +...+ $10000000C7$ ------------------------------------ ( because $nCn$ + $nCr-1$ = $n+1Cr$ )

= $9999999C8$ + $9999999C7$ + $10000000C7$ ------------------------------------ ( because $nCn$ + $nCr-1$ = $n+1Cr$ )

= $10000000C8$ + $10000000C7$ ------------------------------------ ( because $nCn$ + $nCr-1$ = $n+1Cr$ )

= $10000001C8$ ------------------------------------ ( because $nCn$ + $nCr-1$ = $n+1Cr$ )