If S n = r = 1 ∑ n cot − 1 ( r 2 + 3 r + 3 ) , then which of the following options is incorrect ?
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
From the first few n , we note that S n = cot − 1 ( n 2 n + 5 ) . Let us prove by induction that the claim is true for all n ≥ 1 .
Proof: For n = 1 , S 1 = cot − 1 ( 1 2 + 3 ( 1 ) + 3 ) = cot − 1 ( 1 2 ( 1 ) + 5 ) = 7 . Therefore, the claim is true for n = 1 . Assuming the claim is true for n , then:
S n + 1 = S n + cot − 1 ( ( n + 1 ) 2 + 3 ( n + 1 ) + 3 ) = cot − 1 ( n 2 n + 5 ) + cot − 1 ( n 2 + 5 n + 7 ) = cot − 1 ( n 2 n + 5 + n 2 + 5 n + 7 ( n 2 n + 5 ) ( n 2 + 5 n + 7 ) − 1 ) = cot − 1 ( n 3 + 5 n 2 + 9 n + 5 2 n 3 + 1 5 n 2 + 3 8 n + 3 5 ) = cot − 1 ( ( n + 1 ) ( n 2 + 4 n + 5 ) ( 2 n + 7 ) ( n 2 + 4 n + 5 ) ) = cot − 1 ( n + 1 2 ( n + 1 ) + 5 )
Therefore, the claim is also true for n + 1 and hence true for all n ≥ 1 .
And we have:
S 5 S 6 S 8 S ∞ = 5 2 ( 5 ) + 5 = 3 = 6 2 ( 6 ) + 5 = 6 1 7 = 8 2 ( 8 ) + 5 = 8 2 1 = 5 = n → ∞ lim n 2 n + 5 = n → ∞ lim 2 + n 5 = 2 True True False True
Problem Loading...
Note Loading...
Set Loading...
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