If \[{{u}_{n}}=\int_{0}^{\pi /4}{{{\tan }^{n}}\theta }d\theta \] then \[{{u}_{n}}+{{u}_{n-2}}\] is:

If \[{{u}_{n}}=\int_{0}^{\pi /4}{{{\tan }^{n}}\theta }d\theta \] then

1.	If \[{{u}_{n}}=\int_{0}^{\pi /4}{{{\tan }^{n}}\theta }d\theta \] then \[{{u}_{n}}+{{u}_{n-2}}\] is:
A.	\[\frac{1}{n-1}\]
B.	\[\frac{1}{n+1}\]
C.	\[\frac{1}{2n-1}\]
D.	\[\frac{1}{2n+1}\]
Answer» B. \[\frac{1}{n+1}\]

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If \[{{u}_{n}}=\int_{0}^{\pi /4}{{{\tan }^{n}}\theta }d\theta \] then \[{{u}_{n}}+{{u}_{n-2}}\] is:

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