What is (-1)^{n+1}(1+1/(n^2)) ?

The solution to (-1)^{n+1}(1+1/(n^2)) is ((-1)^{n+1}(n^2+1))/(n^2)

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simplify (-1)^{n+1}(1+1/(n^2))

simplify

(- 1)^{n + 1} (1 + \frac{1}{n ^{2}})

\frac{( - 1 ) ^{n + 1} ( n ^{2} + 1 )}{n ^{2}}

Solution steps

(- 1)^{n + 1} (1 + \frac{1}{n ^{2}})

Join

1 + \frac{1}{n ^{2}} : \frac{n ^{2} + 1}{n ^{2}}

= (- 1)^{n + 1} \frac{n ^{2} + 1}{n ^{2}}

Apply the fraction rule:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{( - 1 ) ^{n + 1} ( n ^{2} + 1 )}{n ^{2}}

(z - \frac{1}{z})^{2}

\frac{k ( 2 k + 1 ) ( 4 k + 1 )}{3} + (2 (k + 1))^{2}

(x - \frac{5}{4})^{2} + 3

(x sin (θ) + y cos (θ))^{2}

(\frac{( - 1 - 41 )}{2})^{4}

What is (-1)^{n+1}(1+1/(n^2)) ?
The solution to (-1)^{n+1}(1+1/(n^2)) is ((-1)^{n+1}(n^2+1))/(n^2)