Find the sum to n terms of the series (1/1⋅3) +(1/3⋅5)+(1/5⋅7) +...

Question

Find the sum to n terms of the series (1/1⋅3) +(1/3⋅5)+(1/5⋅7) +... (A) (n/(2n-1)) (B) (n/(2n+1)) (C) n+1 (D) n-1

Tardigrade Admin · Accepted Answer

Let Tr be the rth term of the given series. Then, Tr = (1/ (2r-1)(2r+1)), r= 1, 2, 3, ..., n ⇒ Tr = (1/2) ((1/2r-1) - (1/ 2r+1)) ∴ Required sum = ∑ limitsr=1nTr = (1/2)[((1/1)-(1/3))+((1/3)-(1/5))+((1/5)-(1/7))+...+((1/2n-1) -(1/2n+1))] = (1/2)[1-(1/2n+1)] = (n/2n+1)

Q. Find the sum to $n$ terms of the series $\frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + ...$

$\frac{n}{( 2 n - 1 )}$

$\frac{n}{( 2 n + 1 )}$

$n + 1$

$n - 1$

Q. Find the sum to n terms of the series 1⋅31​+3⋅51​+5⋅71​+...

(2n−1)n​

(2n+1)n​

n+1

n−1

Let Tr​ be the rth term of the given series. Then, Tr​=(2r−1)(2r+1)1​,r=1,2,3,...,n ⇒Tr​=21​(2r−11​−2r+11​) ∴ Required sum =r=1∑n​Tr​=21​[(11​−31​)+(31​−51​)+(51​−71​)+...+(2n−11​−2n+11​)] =21​[1−2n+11​]=2n+1n​

Q. Find the sum to $n$ terms of the series $\frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + ...$

$\frac{n}{( 2 n - 1 )}$

$\frac{n}{( 2 n + 1 )}$

$n + 1$

$n - 1$