User:K3ITHK

Ok I think I got it. If $$f(x) = \sum^{\infty}_{n=0}a_{n}x^n$$ find the value of f'(1) $$a_{0} = 1$$ and $$a_{n} = (7/n)a_{n-1}$$

$$f(x) = 1/0! + 7x/1! + 7^2x^2/2! + 7^3x^3/3!... \sum^{\infty}_{n=0}\frac{7^nx^n}{n!} $$ $$f'(x) = 0 + 7/0! + 7^2x/1! + 7^3x^2/2! ... \sum^{\infty}_{n=1}\frac{7^n x^{n-1}}{(n-1)!} = \sum^{\infty}_{n=1}\frac{n7^nx^{n-1}}{n!} $$ $$f'(1) = 7 + 7^2/1! + 7^3/2! ... \sum^{\infty}_{n=1}\frac{7^n }{(n-1)!} = 7(1 + 7 + 7^2/2! + 7^3/3!...) $$ $$e^x = 1 + x + x^2/2! + x^3/3! +x^4/4!... \sum^{\infty}_{n=0}\frac{x^n }{n!} $$ $$e^7 = 1 + 7 + 7^2/2! + 7^3/3! + 7^4/4!... \sum^{\infty}_{n=0}\frac{7^n }{n!}$$ Recall that: $$f'(1) = 7 + 7^2/1! + 7^3/2! ... \sum^{\infty}_{n=1}\frac{7^n }{(n-1)!} = 7(1 + 7 + 7^2/2! + 7^3/3!...)$$ and compare to $$e^7$$, which gives: $$7e^7$$ Therefore $$f'(1) = 7e^7$$

Sorry if that is a bit hard to follow.