sum of terms of series

Question

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?

user · Accepted Answer

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)$$

Z Ahmed · Answer

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

sum of terms of series

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