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$Hubre usteptau sceguecere pega: sum_(1)^(oo)((x+1)^(n))/(2n+1) Belupice 4 rpaeses, 5 wayobur uapa. Burlumavor 3. Hawor befceerivocro, Mro be mafer volyobe 4. Uecuegobare me exeguneerb. {:[" 4. Uecegog "],[sum_(1)^(1)(n^(2)+2n+3)/(4n^(5)+6n-1)],[" 5.) "(el)={x^(2)+y^(3)z;e^(y)+x^(2)z,z^(6)}]:} Hevira dis i(en) (15)$

$1$ . Hubre usteptau sceguecere pega: $\sum_{1}^{\infty} \frac{(x+1)^{n}}{2 n+1}$ $\newline$ $3$ . Belupice $4$ rpaeses, $5$ wayobur uapa. Burlumavor $3$ . Hawor befceerivocro, Mro be mafer volyobe $4$ . Uecuegobare me exeguneerb. $\newline$ $\begin{array}{l} \text { 4. Uecegog } \\ \sum_{1}^{1} \frac{n^{2}+2 n+3}{4 n^{5}+6 n-1} \\ \text { 5.) }(e l)=\left\{x^{2}+y^{3} z ; e^{y}+x^{2} z, z^{6}\right\} \end{array}$ $\newline$ Hevira dis i(en) $\newline$ ( $15$ )

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Math Problems

Calculus

Find derivatives of using multiple formulae

Full solution

1

. Hubre usteptau sceguecere pega:

\sum_{1}^{\infty} \frac{(x+1)^{n}}{2 n+1}

\newline

3

. Belupice

4

rpaeses,

5

wayobur uapa. Burlumavor

3

. Hawor befceerivocro, Mro be mafer volyobe

4

. Uecuegobare me exeguneerb.

\newline

\begin{array}{l} \text { 4. Uecegog } \\ \sum_{1}^{1} \frac{n^{2}+2 n+3}{4 n^{5}+6 n-1} \\ \text { 5.) }(e l)=\left\{x^{2}+y^{3} z ; e^{y}+x^{2} z, z^{6}\right\} \end{array}

\newline

Hevira dis i(en)

\newline

(

15

)

Identify general term: Identify the general term of the series. $\newline$ The general term of the series is $(x+1)^n / (2n+1)$ .
Recognize series type: Recognize the series type. $\newline$ This series does not match common series types (geometric, telescoping, harmonic) directly and does not simplify easily for direct summation.
Consider convergence: Consider convergence. $\newline$ For $|x+1| < 1$ , the series converges absolutely due to the ratio test or comparison test, as the terms $(x+1)^n$ become very small. For $|x+1| \geq 1$ , convergence is not guaranteed.
Express in known form: Attempt to express the series in a known form. This series does not directly match any standard series expansions or simplifications known at basic calculus levels.
Conclude evaluation: Conclude the evaluation. $\newline$ Without advanced techniques or further information, such as the context of a generating function or a special function, we cannot find a closed form for the sum of the series.