Prove that \(\frac{2n+1}{\sqrt{{{n}^{2}}+2n}+\sqrt{{{n}^{2}}-1}}=\sqrt{{{n}^{2}}+2n}-\sqrt{{{n}

Question

Prove that $\frac{2n+1}{\sqrt{{{n}^{2}}+2n}+\sqrt{{{n}^{2}}-1}}=\sqrt{{{n}^{2}}+2n}-\sqrt{{{n}^{2}}-1}$.[2] Hence find $\sum\limits_{n=1}^{N}{\frac{2n+1}{\sqrt{{{n}^{2}}+2n}+\sqrt{{{n}^{2}}-1}}}$.[3] Deduce the value of $\sum\limits_{n=2}^{N}{\frac{2n-1}{\sqrt{{{n}^{2}}-2n}+\sqrt{{{n}^{2}}-1}}}$.[3] Show that $\sum\limits_{n=1}^{N}{\frac{2n+1}{2n-1}}>\sqrt{{{N}^{2}}+2N}$.[1]

Accepted Answer

$\sqrt{N^{2}+2N}$ (a) $\sqrt{N^{2}-1}$

2016 HCI P1 Q4

2016 HCI P1 Q4

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