# 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?

With more details, we have that

$$frac{2n+sqrt{(4n^2-1)}}{sqrt{2n+1}+sqrt{2n-1}}=frac12left({sqrt{2n+1}+sqrt{2n-1}}right)$$

then

$$frac{4n+sqrt{(4n^2-1)}}{sqrt{2n+1}+sqrt{2n-1}}=frac12left({sqrt{2n+1}+sqrt{2n-1}}right)+frac{2n}{sqrt{2n+1}+sqrt{2n-1}}=\=frac12left({sqrt{2n+1}+sqrt{2n-1}}right)+nleft({sqrt{2n+1}-sqrt{2n-1}}right)=\=frac12(2n+1)sqrt{2n+1}-frac12(2n-1)sqrt{2n-1}=\$$

$$=frac12left(sqrt{(2n+1)^3}-sqrt{(2n-1)^3}right)$$

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