The power series is $\displaystyle f(x)=\sum \limits _{n=0}^\infty (n+1)x^n$. Is this question asking me to take the derivative of the function, which I did and found to be $n(n+1)x^{n-1}$, and then to insert values of $n$ until I get an $ax+b$? I did that for $n=1$ and $n=2$ and got $2$ and $6x$, so $a$ and $b$ are $6$ and $2$ but apparently $a$ is meant to be $2$ and $b$ is $1$. Not sure how to get this.
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You need to be careful, but I'm going to be reckless. Notice $f'(x) = \sum_{n=1}^\infty n(n+1) x^{n-1}$. Plug in $x=0$ to get $f'(0)= 2$ (the only case where you don't have an $x$ is $n=1$). Plugging in $0$ to $f$, you get $f(0)=1$ (the only case without an $x$ is $n=0$). Now what's the tangent line for a general differentiable function at $x=0$ given by? – User203940 Jun 16 '22 at 22:59
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It may be a bit less confusing to explicitly write out the first few terms.
$$f(x) = 1 + 2x + 3x^2 + 4x^3 + \sum_{n=4}^\infty(n+1)x^n$$ $$f'(x) = 2 + 6x + 12x^2 + \sum_{n=4}^\infty(n+1)nx^{n-1}$$
Now, just plug in $x = 0$. All the terms with a positive power of $x$ disappear, leaving:
$$f(0) = 1$$ $$f'(0) = 2$$
Recall that the tangent line at $x = c$ is $y = f'(c)(x - c) + f(c)$. So at $x = 0$, $y = f'(0)x + f(0)$. So for the given $f$, $y = 2x + 1$.
Dan
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