Question

QUIZZZ

[tex]\displaystyle\sum_{k=1}^{50}\left(\sqrt[n-1]{\frac{ {k}^{1 - n} +(k + 1 {)}^{1 - n} }{{k}^{n - 1} +(k + 1 {)}^{ n - 1}}}\right) = [/tex]

A.[tex]\displaystyle1+\frac{1}{51}[/tex]

B.[tex]\displaystyle1-\frac{2}{51}[/tex]

C.[tex]\displaystyle\frac{50}{51}[/tex]

D.[tex]\displaystyle1\frac{1}{51}[/tex]

#backtoschool
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Answer 1

Verified answer

[tex]\begin{aligned}\sum^{50}_{k=1}\left(\sqrt[n-1]{\frac{k^{1-n}+(k+1)^{1-n}}{k^{n-1}+(k+1)^{n-1}}}\right)&=\sum^{50}_{k=1}\left(\sqrt[n-1]{\frac{\frac{1}{k^{n-1}}+\frac{1}{(k+1)^{n-1}}}{k^{n-1}+(k+1)^{n-1}}}\right) \\ &=\sum^{50}_{k=1}\left(\sqrt[n-1]{\frac{\frac{(k+1)^{n-1}+k^{n-1}}{(k^{n-1})((k+1)^{n-1})}}{k^{n-1}+(k+1)^{n-1}}}\right) \\ &=\sum^{50}_{k=1}\left(\sqrt[n-1]{\frac{1}{(k^{n-1})((k+1)^{n-1})}}\right)\end{aligned}[/tex]

[tex]\begin{aligned}\sum^{50}_{k=1}\left(\sqrt[n-1]{\frac{1}{(k^{n-1})((k+1)^{n-1})}}\right)&=\sum^{50}_{k=1}\left(\sqrt[n-1]{(k(k+1))^{-(n-1)}}\right) \\ &=\sum^{50}_{k=1}\left(k(k+1)\right)^{-1} \\ &=\sum^{50}_{k=1}\frac{1}{k(k+1)} \\ &=\sum^{50}_{k=1}\frac{(k+1)-k}{k(k+1)} \\ &=\sum^{50}_{k=1}\frac{1}{k}-\frac{1}{k+1} \\ &=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\dots+\left(\frac{1}{50}-\frac{1}{51}\right) \\ &=1-\frac{1}{51} \\ &=\frac{50}{51}\end{aligned}[/tex]