Question

Next \infty 2n If $C(x) = \sum_{n=0} \frac{2n}{4n^2 + 1} x^{2n}$ and $S(x) = \sum_{n=0} \frac{2n+1}{(2n+1)^2 + 1} x^{2n+1}$, find the power series of $C(x) - S(x)$. $\infty$ $\sum_{n=0}$

          Next
\infty
2n
If $C(x) = \sum_{n=0} \frac{2n}{4n^2 + 1} x^{2n}$ and $S(x) = \sum_{n=0} \frac{2n+1}{(2n+1)^2 + 1} x^{2n+1}$, find the power series of
$C(x) - S(x)$.
$\infty$
$\sum_{n=0}$

Next
∞2n
If C(x) = ∑n=0(2n)/(4n^2 + 1) x^2n and S(x) = ∑n=0(2n+1)/((2n+1)^2 + 1) x^2n+1, find the power series of
C(x) - S(x).
∞
∑n=0

Added by Vincent V.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Krishna G

Step 1

We can rewrite C as 2n^2 and S as 3^(2n+1) / (2n+1)^2: (4n+1) / (n^2+1) * Cx - Sx = (4n+1) / (n^2+1) * (2n^2)x - (3^(2n+1) / (2n+1)^2)x Now, let's simplify further: = (8n^3 + 2n) / (n^2+1) * x - (3^(2n+1) / (2n+1)^2)x Show more…

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