Use partial fractions to find the power series of the function: (9x^(2)+184)/((x^(2)+16)(x^(2)+36) sum_(n=0)^(oo)

Express Function: Step 1: Express the function using partial fractions. We need to decompose the function (9x^2 + 184)/((x^2 + 16)(x^2 + 36)) into simpler fractions. Assume (9x^2 + 184)/((x^2 + 16)(x^2 + 36)) = A/(x^2 + 16) + B/(x^2 + 36). Multiply through by (x^2 + 16)(x^2 + 36) to clear the denominators: 9x^2 + 184 = A(x^2 + 36) + B(x^2 + 16).Solve for A and B: Step 2: Solve for A and B. Equating coefficients, we get: For x^2: 9 = A + B For constant terms: 184 = 36A + 16B Solving these equations: 9 = A + B (1) 184 = 36A + 16B (2) From (1), B = 9 - A. Substitute into (2): 184 = 36A + 16(9 - A) 184 = 36A + 144 - 16A 20A = 40 A = 2, B = 7.Write Decomposition: Step 3: Write the partial fraction decomposition. Substitute values of A and B back: (9x^2 + 184)/((x^2 + 16)(x^2 + 36)) = 2/(x^2 + 16) + 7/(x^2 + 36).Expand Power Series: Step 4: Expand each term into a power series. For 2/(x^2 + 16), use the geometric series formula: 2/(x^2 + 16) = 2 * 1/(16(1 + (x^2/16))) = (1/8) * sum_(n=0)^(oo) ((-1)^n (x^2/16)^n) = sum_(n=0)^(oo) ((-1)^n x^(2n)/128^n). For 7/(x^2 + 36), similarly: 7/(x^2 + 36) = 7 * 1/(36(1 + (x^2/36))) = (7/36) * sum_(n=0)^(oo) ((-1)^n (x^2/36)^n) = sum_(n=0)^(oo) ((-1)^n 7x^(2n)/1296^n).

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Math Problems

Algebra 2

Sum of finite series not start from 1

Full solution

Q. Use partial fractions to find the power series of the function:

\frac{9 x^{2}+184}{\left(x^{2}+16\right)\left(x^{2}+36)\right.}

\newline

\sum_{n=0}^{\infty}

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Express Function: Step $1$ : Express the function using partial fractions. $\newline$ We need to decompose the function $(9x^2 + 184)/((x^2 + 16)(x^2 + 36))$ into simpler fractions. $\newline$ Assume $(9x^2 + 184)/((x^2 + 16)(x^2 + 36)) = A/(x^2 + 16) + B/(x^2 + 36)$ . $\newline$ Multiply through by $(x^2 + 16)(x^2 + 36)$ to clear the denominators: $\newline$ $9x^2 + 184 = A(x^2 + 36) + B(x^2 + 16)$ .
Solve for A and B: Step $2$ : Solve for A and B. $\newline$ Equating coefficients, we get: $\newline$ For $x^2$ : $9 = A + B$ $\newline$ For constant terms: $184 = 36A + 16B$ $\newline$ Solving these equations: $\newline$ $9 = A + B$ ( $1$ ) $\newline$ $184 = 36A + 16B$ ( $2$ ) $\newline$ From ( $1$ ), $B = 9 - A$ . $\newline$ Substitute into ( $2$ ): $\newline$ $184 = 36A + 16(9 - A)$ $\newline$ $184 = 36A + 144 - 16A$ $\newline$ $20A = 40$ $\newline$ $A = 2$ , $9 = A + B$ $0$ .
Write Decomposition: Step $3$ : Write the partial fraction decomposition. $\newline$ Substitute values of $A$ and $B$ back: $\newline$ $(9x^2 + 184)/((x^2 + 16)(x^2 + 36)) = \frac{2}{x^2 + 16} + \frac{7}{x^2 + 36}$ .
Expand Power Series: Step $4$ : Expand each term into a power series. $\newline$ For $\frac{2}{x^2 + 16}$ , use the geometric series formula: $\newline$ $\frac{2}{x^2 + 16} = 2 \cdot \frac{1}{16(1 + \frac{x^2}{16})}$ $\newline$ $= \frac{1}{8} \cdot \sum_{n=0}^{\infty} ((-1)^n (\frac{x^2}{16})^n)$ $\newline$ $= \sum_{n=0}^{\infty} ((-1)^n \frac{x^{2n}}{128^n})$ . $\newline$ For $\frac{7}{x^2 + 36}$ , similarly: $\newline$ $\frac{7}{x^2 + 36} = 7 \cdot \frac{1}{36(1 + \frac{x^2}{36})}$ $\newline$ $= \frac{7}{36} \cdot \sum_{n=0}^{\infty} ((-1)^n (\frac{x^2}{36})^n)$ $\newline$ $= \sum_{n=0}^{\infty} ((-1)^n \frac{7x^{2n}}{1296^n})$ .