sum of terms of series

Mathematics Asked by Albus Dumbledore on December 2, 2020

If $$F(t)=displaystylesum_{n=1}^tfrac{4n+sqrt{4n^2-1}}{sqrt{2n+1}+sqrt{2n-1}}$$ find $F(60)$.

I tried manipulating the general term(of sequence) in the form $V(n)-V(n-1)$ to calculate the sum by cancellation but went nowhere. I also tried using the fact that $$2n+sqrt{4n^2-1}=frac{1}{2}{(sqrt{2n-1}+sqrt{2n+1})}^2$$ Could someone please give me a hint?

2 Answers

With more details, we have that





Correct answer by user on December 2, 2020

$$T_n=frac{4n+sqrt{4n^2-1}}{sqrt{2n+1}+sqrt{2n-1}}=frac{1}{2}[2n+1)^{3/2}-(2n-1)^{3/2}]=$$ Next by Telescopic summing we get $$S(60)=[121^{3/2}-1]/2=[1331-1]/2=665$$

Answered by Z Ahmed on December 2, 2020

Add your own answers!

Related Questions

Geometric proof for the half angle tangent

3  Asked on November 12, 2021 by brazilian_student


Skyscrapers sheaf’s global sections

2  Asked on November 12, 2021 by abramo


Ideals in a UFD

1  Asked on November 12, 2021


Solving this DE

1  Asked on November 12, 2021


Prove that $a ⊈ {a}$, where $a$ is non-empty

1  Asked on November 12, 2021 by galaxylokka


Ask a Question

Get help from others!

© 2022 All rights reserved. Sites we Love: PCI Database, MenuIva, UKBizDB, Menu Kuliner, Sharing RPP, SolveDir