Integrated Progressions

Calculus Level 4

let $a_{n} =\displaystyle \int_{0}^{\pi/4}\tan^{n}x \mathrm{d}x$ . Then $a_{2} + a_{4}, a_{3}+a_{5},a_{4}+a_{6}$ are in what kind of progression?

Arithmetic progression Harmonic progression Geometric progression None of these choices

1 solution

Vishwak Srinivasan
Apr 19, 2015

Using reduction formula, it is easy to find out that

$a_{n} + a_{n-2} = \dfrac{1}{n-1}$

$a_{4} + a_{2} = \dfrac{1}{3}$

$a_{5} + a_{3} = \dfrac{1}{4}$

$a_{6} + a_{4} = \dfrac{1}{5}$

$\dfrac{1}{3} , \dfrac{1}{4} , \dfrac{1}{5}$ are in H.P

Proof of $a_{n} + a_{n-2} = \dfrac{1}{n-1}$

$a_n = \displaystyle \int_{0}^{\pi/4} \tan^{n-2}x \times \tan^2x dx$

$a_n = \displaystyle \int_{0}^{\pi/4} \tan^{n-2}x \times (\sec^2x - 1)dx$

$a_n = \displaystyle \int_{0}^{\pi/4} \tan^{n-2}x \times \sec^2x dx - \int_{0}^{\pi/4} \tan^{n-2}x dx$

$a_n + a_{n-2} = \displaystyle \int_{0}^{\pi/4} \tan^{n-2}x \times \sec^2x dx$

Take $\tan x = t$

$a_n + a_{n-2} = \displaystyle \int_{0}^{1} t^{n-2} dt$

$a_n + a_{n-2} = \boxed{\dfrac{1}{n-1}}$

This solution has been marked incomplete. It would be better to show how you obtain $a_n + a_{n-2} = \frac 1 {n-1}$ .

I have now included the proof of $a_n + a_{n-2} = \dfrac{1}{n-1}$ .

Vishwak Srinivasan - 6 years, 1 month ago