Math, asked by sajan6491, 18 days ago

 \pink{ \rm{For  \: each  \: positive \:  integer \:  n, let}} \\  \rm \color{blue}f(n) =   \frac{1}{ \sqrt[3]{ {n}^{2} + 2n + 1 }  +  \sqrt[3]{ {n}^{2} - 1 } +  \sqrt[3]{ {n}^{2}  - 2n + 1}  }    \\  \color{lime} \rm{determine \: the \: value \: of} \\  \color{darkgreen} \rm f(1) + f(3) + f(5) +  \dots \dots + f(999997) + f(999999)

Answers

Answered by mathdude500
5

\large\underline{\sf{Solution-}}

Given expression is

\rm \: f(n) = \dfrac{1}{ \sqrt[3]{ {n}^{2} + 2n + 1 } + \sqrt[3]{ {n}^{2} - 1 } + \sqrt[3]{ {n}^{2} - 2n + 1} }

can be rewritten as

\rm \: f(n) = \dfrac{1}{ \sqrt[3]{ {(n + 1)}^{2} } + \sqrt[3]{(n + 1)(n - 1)} + \sqrt[3]{ {(n - 1)}^{2}} }

can be further rewritten as

\rm \: f(n) = \dfrac{1}{ {( \sqrt[3]{n + 1}) }^{2}  +  \sqrt[3]{n + 1} \sqrt[3]{n - 1}  +  {( \sqrt[3]{n - 1} )}^{2} }  \\

We know,

\boxed{ \rm{ \: \frac{ {x}^{3}  -  {y}^{3} }{x - y} =  {x}^{2} + xy +  {y}^{2} \: }} \\

can also be rewritten as

\color{green}\boxed{ \rm{ \: \frac{x - y}{ {x}^{3}  -  {y}^{3} } =  \frac{1}{ {x}^{2}  + xy +  {y}^{2} } \: }} \\

So, using this identity, here

\rm \: x =  \sqrt[3]{n + 1}  \\

and

\rm \: y =  \sqrt[3]{n  -  1}  \\

So, on applying the result, we get

\rm \: f(n) = \dfrac{ \sqrt[3]{n + 1}  -  \sqrt[3]{n - 1} }{n + 1 - (n - 1)}

\rm \: f(n) = \dfrac{ \sqrt[3]{n + 1}  -  \sqrt[3]{n - 1} }{n + 1 - n + 1}

\rm \: f(n) = \dfrac{ \sqrt[3]{n + 1}  -  \sqrt[3]{n - 1} }{2}  \\

So, on substituting n = 1, 3, 5, ..., 999999, we get

\rm \: f(1) = \dfrac{ \sqrt[3]{2}}{2}  \\

\rm \: f(3) = \dfrac{ \sqrt[3]{4}  -  \sqrt[3]{2} }{2}  \\

\rm \: f(5) = \dfrac{ \sqrt[3]{6}  -  \sqrt[4]{2} }{2}  \\

\rm \: f(7) = \dfrac{ \sqrt[3]{8}  -  \sqrt[6]{2} }{2}  \\

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\rm \: f(999999) = \dfrac{ \sqrt[3]{1000000}  -  \sqrt[3]{999998} }{2}  \\

So, on adding above all we get

\rm \: f(1) + f(3) + f(5) + f(7) +  \cdots + f(999999) \\

\rm \: = \dfrac{ \sqrt[3]{2} }{2} + \dfrac{ \sqrt[3]{4}  -  \sqrt[3]{2} }{2} +\dfrac{ \sqrt[3]{6}  -  \sqrt[3]{4} }{2} +  \cdots + \dfrac{ \sqrt[3]{1000000}  -  \sqrt[3]{999998} }{2}   \\

\rm \:  =  \: \dfrac{1}{2}\bigg( \sqrt[3]{2}  +  \sqrt[3]{4} -  \sqrt[3]{2} +  \sqrt[3]{6} -  \sqrt[3]{4} +  \cdots +  \sqrt[3]{1000000} -  \sqrt[3]{999998}\bigg)

\rm \:  =  \: \dfrac{1}{2}\bigg( \sqrt[3]{1000000} \bigg)

\rm \:  =  \: \dfrac{1}{2} \times 100

\rm \:  =  \: 50 \\

Hence,

{ \rm{For \: each \: positive \: integer \: n, if}} \\ \rm \color{blue}f(n) = \frac{1}{ \sqrt[3]{ {n}^{2} + 2n + 1 } + \sqrt[3]{ {n}^{2} - 1 } + \sqrt[3]{ {n}^{2} - 2n + 1} } \\ \color{red} \rm{ \: then \: the \:  \: value \: of} \\ \color{green} \rm f(1) + f(3) + f(5) + \dots \dots + f(999997) + f(999999) = 50 \\

\rule{190pt}{2pt}

Additional Information :-

\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: identities}}}} \\ \\ \bigstar \: \bf{ {(x + y)}^{2} =  {x}^{2}  + 2xy +  {y}^{2} }\:\\ \\ \bigstar \: \bf{ {(x - y)}^{2}  =  {x}^{2} - 2xy +  {y}^{2} }\:\\ \\ \bigstar \: \bf{ {x}^{2} -  {y}^{2} = (x + y)(x - y)}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2}  -  {(x - y)}^{2}  = 4xy}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2}  +  {(x - y)}^{2}  = 2( {x}^{2}  +  {y}^{2})}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{3} =  {x}^{3} +  {y}^{3} + 3xy(x + y)}\:\\ \\ \bigstar \: \bf{ {(x - y)}^{3} =  {x}^{3} -  {y}^{3} - 3xy(x - y) }\:\\ \\ \bigstar \: \bf{ {x}^{3}  +  {y}^{3} = (x + y)( {x}^{2}  - xy +  {y}^{2} )}\: \end{array} }}\end{gathered}\end{gathered}\end{gathered}

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