Solve n(n-1)/n^3-3n-5 | Microsoft Math Solver

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\frac{n^{2}-n}{n^{3}-3n-5}

Short Solution Steps

\frac { n ( n - 1 ) } { n ^ { 3 } - 3 n - 5 }

Use the distributive property to multiply n by n-1.

\frac{n^{2}-n}{n^{3}-3n-5}

Evaluate

\frac{n\left(n-1\right)}{n^{3}-3n-5}

Short Solution Steps

\frac { n ( n - 1 ) } { n ^ { 3 } - 3 n - 5 }

Use the distributive property to multiply n by n-1.

\frac{n^{2}-n}{n^{3}-3n-5}

Quiz

Polynomial

5 problems similar to:

\frac { n ( n - 1 ) } { n ^ { 3 } - 3 n - 5 }

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\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}} ...

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f(n) = [(n + 1)^(n-1)]/n^n We ignore the uninteresting case that n = 0. So f(1) = 1 which is = 1: marginal. f(2) = 3/4 < 1 ; ok, now we can start. If f(n) < 1 . then f(n+1) = [(n + ...

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Write \frac{1}{n^2+n} = \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}. See if now you can identify a telescoping sequence as follows : \sum_{n=1}^{2008} \frac{1}{n^2+n} = \sum_{n=1}^{2008} \left(\frac{1}{n} - \frac{1}{n+1}\right) = 1 - \frac{1}{2009} ...

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\frac{n^{2}-n}{n^{3}-3n-5}

Use the distributive property to multiply n by n-1.

\frac{n^{2}-n}{n^{3}-3n-5}

Use the distributive property to multiply n by n-1.

Examples

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{ x } ^ { 2 } - 4 x - 5 = 0

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\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right]

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\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.

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\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }

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