sum of terms of series

Question

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?

Your sum seems to be ill defined, since no indexing terms seem to appear in the summand. Do you mean to have $i$ instead of $n$ in the summands? — spaceman, Jul 31 '20 at 14:40
I would multiply with $$\frac{\sqrt{2n+1} - \sqrt{2n-1}}{\sqrt{2n+1} - \sqrt{2n-1}}$$ for a start. — Daniel Fischer, Jul 31 '20 at 14:49

score 1 · Answer 1 · answered Jul 31 '20 at 15:01

1

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

answered Jul 31 '20 at 15:01

Z Ahmed

43,235

score 1 · Accepted Answer · answered Jul 31 '20 at 15:07

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

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

$$=\frac12\left(\sqrt{(2n+1)^3}-\sqrt{(2n-1)^3}\right)$$

sum of terms of series

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