यदि {(1 + x)^{14}} के विस्तार में {T_r} | vedclass

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Question

(d) ${T_{r\,}} = {\,^{14}}{C_{r - 1}}\,{x^{r - 1}}\,;\,\,{T_{r + 1}}\, = \,{\,^{14}}{C_r}\,{x^r}$; ${T_{r + 2}} = {\,^{14}}{C_{r + 1}}\,{x^{r + 1}}$

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(d) ${T_{r\,}} = {\,^{14}}{C_{r - 1}}\,{x^{r - 1}}\,;\,\,{T_{r + 1}}\, = \,{\,^{14}}{C_r}\,{x^r}$; ${T_{r + 2}} = {\,^{14}}{C_{r + 1}}\,{x^{r + 1}}$

By the given condition $2\,{.^{14}}{C_r} = {\,^{14}}{C_{r - 1}} + {\,^{14}}{C_{r + 1}}$ &hellip;..$(i)$

==> $2\,.\frac{{14!}}{{r!\,\,(14 - r)!}}\, = \,\frac{{14!}}{{(r - 1)!\,\,(15 - r)!}} + \frac{{14!}}{{(r + 1)!\,(13 - r)!}}$

==> $\frac{2}{{r.(r - 1)!.(14 - r).(13 - r)!}}$=$\frac{1}{{(r - 1)!.(15 - r).(14 - r).(13 - r)!}}$

+$\frac{1}{{(r + 1)\,r(r - 1)\,!\,(13 - r)\,!}}$

==> $\frac{2}{{r(14 - r)}} = \frac{1}{{(15 - r)(14 - r)}} + \frac{1}{{(r + 1)r}}$

==> $\frac{1}{{r(14 - r)}} - \frac{1}{{(15 - r)(14 - r)}} = \frac{1}{{(r + 1)r}} - \frac{1}{{r(14 - r)}}$

==> $\frac{{(15 - r) - r}}{{r(15 - r)(14 - r)}} = \frac{{(14 - r) - (r + 1)}}{{(r + 1)r(14 - r)}}$

==> $\frac{{15 - 2r}}{{15 - r}} = \frac{{13 - 2r}}{{r + 1}}$

==> $15r + 15 - 2{r^2} - 2r = 195 - 30r - 13r + 2{r^2}$

==> $4{r^2} - 56r + 180 = 0\,\,$

$\Rightarrow \,{r^2} - 14r + 45 = 0$

==> $(r - 5)(r - 9) = 0 $

$\Rightarrow r = 5,\,9$

But $5$ is not given.

Hence $r = 9.$ 

$Trick :$ Put the value of $r$ from options in equation $(i)$, only $(d)$ satisfy it.

यदि ${(1 + x)^{14}}$ के विस्तार में ${T_r},\,{T_{r + 1}},\,{T_{r + 2}}$ के गुणांक समांतर श्रेणी में हों, तो $r = $

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