Inverse Trigonometric Functions#1

Calculus Level 3

If S n = r = 1 n cot 1 ( r 2 + 3 r + 3 ) S_n = \displaystyle \sum_{r=1}^n \cot^{-1} (r^2+3r+3) , then which of the following options is incorrect ?

S = cot 1 2 S_\infty = \cot^{-1} 2 S 6 = cot 1 17 6 S_6 = \cot^{-1} \frac {17}6 S 5 = cot 1 3 S_5 = \cot^{-1} 3 S 8 = cot 1 5 S_8 = \cot^{-1} 5

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2 solutions

Ashutosh Sharma
Mar 4, 2018

arccot[(r^2+3r+2)+1]=arccot[(r+2)(r+1)+1] which is terminating series now as shown below therefore Sn= arccot(r+1)-arccot(r+2).rest is simple

Chew-Seong Cheong
Feb 26, 2018

From the first few n n , we note that S n = cot 1 ( 2 n + 5 n ) S_n = \cot^{-1} \left(\dfrac {2n+5}n\right) . Let us prove by induction that the claim is true for all n 1 n\ge 1 .

Proof: For n = 1 n=1 , S 1 = cot 1 ( 1 2 + 3 ( 1 ) + 3 ) = cot 1 ( 2 ( 1 ) + 5 1 ) = 7 S_1 = \cot^{-1}\left(1^2+3(1)+3\right) = \cot^{-1} \left(\dfrac {2(1)+5}1\right) = 7 . Therefore, the claim is true for n = 1 n=1 . Assuming the claim is true for n n , then:

S n + 1 = S n + cot 1 ( ( n + 1 ) 2 + 3 ( n + 1 ) + 3 ) = cot 1 ( 2 n + 5 n ) + cot 1 ( n 2 + 5 n + 7 ) = cot 1 ( ( 2 n + 5 n ) ( n 2 + 5 n + 7 ) 1 2 n + 5 n + n 2 + 5 n + 7 ) = cot 1 ( 2 n 3 + 15 n 2 + 38 n + 35 n 3 + 5 n 2 + 9 n + 5 ) = cot 1 ( ( 2 n + 7 ) ( n 2 + 4 n + 5 ) ( n + 1 ) ( n 2 + 4 n + 5 ) ) = cot 1 ( 2 ( n + 1 ) + 5 n + 1 ) \begin{aligned} S_{n+1} & = S_n + \cot^{-1} \left((n+1)^2 + 3(n+1)+3\right) \\ & = \cot^{-1} \left(\frac {2n+5}n \right) + \cot^{-1} \left(n^2 + 5n+7\right) \\ & = \cot^{-1} \left(\frac {\left(\frac {2n+5}n \right)\left(n^2 + 5n+7\right)-1}{\frac {2n+5}n + n^2 + 5n+7} \right) \\ & = \cot^{-1} \left(\frac {2n^3+15n^2+38n+35}{n^3+5n^2+9n+5} \right) \\ & = \cot^{-1} \left(\frac {(2n+7)(n^2+4n+5)}{(n+1)(n^2+4n+5)} \right) \\ & = \cot^{-1} \left(\frac {2(n+1)+5}{n+1} \right) \end{aligned}

Therefore, the claim is also true for n + 1 n+1 and hence true for all n 1 n \ge 1 .

And we have:

S 5 = 2 ( 5 ) + 5 5 = 3 True S 6 = 2 ( 6 ) + 5 6 = 17 6 True S 8 = 2 ( 8 ) + 5 8 = 21 8 5 False S = lim n 2 n + 5 n = lim n 2 + 5 n = 2 True \begin{aligned} S_5 & = \frac {2(5)+5}5 = \color{#3D99F6} 3 & \color{#3D99F6} \text{True} \\ S_6 & = \frac {2(6)+5}6 = \color{#3D99F6} \frac {17}6 & \color{#3D99F6} \text{True} \\ S_8 & = \frac {2(8)+5}8 = \color{#D61F06} \frac {21}8 \ne 5 & \color{#D61F06} \text{False} \\ S_\infty & = \lim_{n\to \infty} \frac {2n+5}n = \lim_{n\to \infty} 2 + \frac 5n = \color{#3D99F6} 2 & \color{#3D99F6} \text{True} \end{aligned}

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