Math, asked by Anonymous, 1 year ago

Find {(p+2)}^{th} term from the end in
 {(x -  \frac{1}{x} )}^{2n + 1}

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Answers

Answered by rahman786khalilu
2

Given, ( x + 1/x)^2n+1

Number of the terms = 2n + 2

(p + 2 ) terms from the last =

(2n + 2) - (p+ 2) +1

=(2n - p + 1)

term from starting

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Answered by generalRd
7

ANSWER

(-1)^{2n-p}.{}^{2n+1}C_{2n-p}(x)^{2p -2n +1}

Step By Step Explanation

Here, we have →

The expansion of (x - \dfrac{1} {x})^{2n + 1} has (2n+1) terms.

So,(p + 2)^{th} term from end →

[(2n + 2) - (p +2) + 1]^{th} term from beginning

i.e (2n-p+1)^{th} term.

Now, we have →

T_{2n-p+1} ={ }^{2n+1}C_{2n-p}(x)^{p + 1}(\dfrac{-1}{x})^{2n-p}

\implies T_{2n-p+1} = (-1)^{2n-p}.{}^{2n+1}C_{2n-p}(x)^{2p -2n +1}


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