Question

S = `tan^-1(1/(n^2 + n + 1)) + tan^-1(1/(n^2 + 3n + 3)) + ... + tan^-1(1/(1 + (n + 19)(n + 20)))`, then tan S is equal to ______.

Options

• `20/(401 + 20n)`

• `n/(n^2 + 20n + 1)`

• `20/(n^2 + 20n + 1)`

• `n/(401 + 20n)`

Answer 1

S = `tan^-1(1/(n^2 + n + 1)) + tan^-1(1/(n^2 + 3n + 3)) + ... + tan^-1(1/(1 + (n + 19)(n + 20)))`, then tan S is equal to `underlinebb(20/(n^2 + 20n + 1))`.

Explanation:

Given

S = `tan^-1(1/(n^2 + n + 1)) + tan^-1(1/(n^2 + 3n + 3)) + ... + tan^-1(1/(1 + (n + 19)(n + 20)))`

= `tan^-1((n + 1 - n)/(1 + n(n + 1))) + tan^-1[((n + 2) - (n - 1))/(1 + (n + 1)(n + 2))] + ...`

= [tan^–1(n + 1) – tan^–1(n)] + [tan^–1(n + 2) – tan^–1(n + 1)] + ...... + [tan^–1(n + 20) – tan^–1(n + 19)]

`\implies` tan^–1(n + 20) – tan^–1(n)

= `tan^-1(20/(1 + n^2 + 20n))` or tan 8 = `20/(n^2 + 20n + 1)`