<s>[INST] sum of terms of series 
If $$F(t)=\displaystyle\sum_{n=1}^t\frac{4n+\sqrt{4n^2-1}}{\sqrt{2n+1}+\sqrt{2n-1}}$$ 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?
 [/INST] With more details, we have that
$$\frac{2n+\sqrt{(4n^2-1)}}{\sqrt{2n+1}+\sqrt{2n-1}}=\frac12\left({\sqrt{2n+1}+\sqrt{2n-1}}\right)$$
then
$$\frac{4n+\sqrt{(4n^2-1)}}{\sqrt{2n+1}+\sqrt{2n-1}}=\frac12\left({\sqrt{2n+1}+\sqrt{2n-1}}\right)+\frac{2n}{\sqrt{2n+1}+\sqrt{2n-1}}=\\=\frac12\left({\sqrt{2n+1}+\sqrt{2n-1}}\right)+n\left({\sqrt{2n+1}-\sqrt{2n-1}}\right)=\\=\frac12(2n+1)\sqrt{2n+1}-\frac12(2n-1)\sqrt{2n-1}=\\$$
$$=\frac12\left(\sqrt{(2n+1)^3}-\sqrt{(2n-1)^3}\right)$$
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