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

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

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simplify 1/((1+a^{n-1))}+1/((1+a^{m-1))}

simplify

\frac{1}{( 1 + a ^{n - 1} )} + \frac{1}{( 1 + a ^{m - 1} )}

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

Solution steps

\frac{1}{1 + a ^{n - 1}} + \frac{1}{1 + a ^{m - 1}}

Least Common Multiplier of

1 + a^{n - 1}, 1 + a^{m - 1} : (a^{n - 1} + 1) (a^{m - 1} + 1)

Adjust Fractions based on the LCM

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

Apply the fraction rule:

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

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

Simplify

a^{m - 1} + 1 + a^{n - 1} + 1 : a^{m - 1} + a^{n - 1} + 2

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

- (x - 9)

\frac{- 4 x}{2 x}

(2 y - 5) \cdot (3 y^{3.4})

cos (x) \cdot (1 + tan^{2} (x))

2 x^{2} + 2 6 x - 27

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