Help with finding the generating function (with a constant )

Question

How do you get the generating function from this formula:

$8(1+x)^{7}$

I have the following formula for $(1+x)^{n}$ :

$n\choose 0$ + $n \choose 1$$x^1$ + $n \choose 2$$x^2$+... +$n \choose n$$x^n$

so shouldn't this mean for this formula it should be:

8$7 \choose 0$$x^0$+8$7 \choose 1$$x^1$+8$7 \choose 2$$x^2$+....+8$7 \choose 7$$x^7$

but this doesn't seem to be equivalent to the book's answer of:

$8 \choose 1$$x^0$+2$8 \choose 2$$x^1$+3$8 \choose 3$$x^2$+....+8$8 \choose 8$$x^7$

Why is the 8 incrementing up instead of being multiplied by each element of the function?

Answer 1

What they have is the same as what you have. $$8\begin{pmatrix} 7 \\ k \end{pmatrix} = (k+1)\begin{pmatrix} 8 \\ k +1 \end{pmatrix}\,.$$

To see this,

$$8\times {7!\over k! (7-k!)} = {8!\over k! (8 - 1 - k)!} = {8!\over k!(8-(k+1))!} $$
$$ = (k+1)\times {8!\over (k+1)!(8 - (k+1))!} = {8!\over k! (8 - (k+1))!} \,.$$

Answer 2

Suppose $n\geq 1$. Note that $$ n\binom{8}{n}=n\cdot\frac{8!}{n!(8-n)!}=n\cdot\frac{8\cdot 7!}{n\cdot(n-1)!\cdot(8-n)!}. $$ The $n$ in the denominator now cancels: $$ n\binom{8}{n}=8\cdot\frac{7!}{(n-1)!(8-n)!}. $$ Finally, because we can write $8-n=7-(n-1)$, we have $$ n\binom{8}{n}=8\cdot\frac{7!}{(n-1)!(7-(n-1))!}=8\binom{7}{n-1}. $$