Show that the expansion of x2-1-x12 does not contian any term involving x-1-

Any term in the expansion of ( x^2 +1/x )^12 is T_N =T_r+1=^nC_rx^n ry^r =^12C_r(x^2)^12 r ( 1/x )^12 =^12C_rx^24 2rx^ 12 =x^12 2r=x^ 1 12 2r= 1 2r=13 r=13/2 r ...

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Byju's Answer

Standard XII

Mathematics

Index of (r+1)th Term from End When Counted from Beginning

Show that the...

Question

Show that the expansion of ${(x^{2} + \frac{1}{x})}^{12}$ does not contian any term involving $x^{- 1} .$

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Solution

Any term in the expansion of ${(x^{2} + \frac{1}{x})}^{12}$ is

$T_{N} = T_{r + 1} =^{n} C_{r} x^{n - r} y^{r}$

$=^{12} C_{r} (x^{2})^{12 - r} {(\frac{1}{x})}^{12}$

$=^{12} C_{r} x^{24 - 2 r} x^{- 12}$

$= x^{12 - 2 r} = x^{- 1}$

12-2r=-1

2r=13

$r = \frac{13}{2}$

r can not be a fraction, therefore there is no term in the expansion of ${(x^{2} + \frac{1}{x})}^{12}$ having the term $x^{- 1} .$

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Show that the expansion of (x2+1x)12 does not contian any term involving x−1.

Any term in the expansion of (x2+1x)12 is TN=Tr+1=nCrxn−ryr =12Cr(x2)12−r(1x)12 =12Crx24−2rx−12 =x12−2r=x−1 12-2r=-1 2r=13 r=132 r can not be a fraction, therefore there is no term in the expansion of (x2+1x)12having the term x−1.

Show that the expansion of ${(x^{2} + \frac{1}{x})}^{12}$ does not contian any term involving $x^{- 1} .$