Answer
$c_0$
$c_0+c_1(x-3)$
$c_0+c_1(x-3)+c_2(x-3)^2$
$c_0+c_1(x-3)+c_2(x-3)^2+c_3(x-3)^3$
Work Step by Step
The general form of a power series is:
$\sum_{k=0}^{\infty} c_k(x-a)^k$
As the series is centered to 3, rewrite the expression for $a=3$:
$\sum_{k=0}^{\infty} c_k (x-3)^k$
Determine the first 4 terms:
$\sum_{k=0}^{0} c_k(x-3)^k=c_0(x-3)^0=c_0$
$\sum_{k=0}^{1} c_k(x-3)^k=c_0(x-3)^0+c_1(x-3)^1$
$=c_0+c_1(x-3)$
$\sum_{k=0}^{2} c_k(x-3)^k=c_0(x-3)^0+c_1(x-3)^1+c_2(x-3)^2$
$=c_0+c_1(x-3)+c_2(x-3)^2$
$\sum_{k=0}^{3} c_k(x-3)^k=c_0(x-3)^0+c_1(x-3)^1+c_2(x-3)^2+c_3(x-3)^3$
$=c_0+c_1(x-3)+c_2(x-3)^2+c_3(x-3)^3$
