First slide

Binomial theorem for positive integral Index

Question

$The value of numerically greatest term in the expansion (5 x - 2 y)^{7} when$ $x = \frac{1}{2}, y = \frac{1}{3}$

Moderate

A

${(\frac{5}{2})}^{7}, (\frac{14}{3}) {(\frac{5}{2})}^{6}$
B

${(\frac{5}{2})}^{7}, (- \frac{14}{3}) {(\frac{5}{2})}^{6}$
C

$- {(\frac{5}{2})}^{7}, (\frac{14}{3}) {(\frac{5}{2})}^{6}$
D

$- {(\frac{5}{2})}^{7}, (- \frac{14}{3}) {(\frac{5}{2})}^{6}$

Solution

${(5 x - 2 y)}^{7} = {(5 x)}^{7} {(1 - \frac{2 y}{5 x})}^{7}$
$Here n = 7, X = - \frac{2 y}{5 x} = \frac{- 2}{5} \cdot \frac{1}{3} \cdot \frac{2}{1} (∵ x = \frac{1}{2}, y = \frac{1}{3})$
   $\begin{array}{l} = \frac{- 4}{15} \\ \Rightarrow |X| = \frac{4}{15} \end{array}$
$Now, \frac{(n + 1) | X |}{| X | + 1} = \frac{(7 + 1) \times \frac{4}{15}}{\frac{4}{15} + 1} = \frac{8 \times \frac{4}{15}}{\frac{19}{5}} = \frac{32}{19} = 1.68 = 1 + 0.68 = I + F$
$∴ Numerically greatest terms are T_{1} and T_{2}$
$T_{r + 1} = 7_{C_{_{r}}} {(5 x)}^{7 - r} {(- 2 y)}^{r}$
$∴ T_{1} = T_{0 + 1} = 7_{C_{_{0}}} {(5 x)}^{7} {(- 2 y)}^{0}$
$= {1.5}^{7} . {(\frac{1}{2})}^{7} .1 (∵ x = \frac{1}{2}, y = \frac{1}{3})$
$= {(\frac{5}{2})}^{7}$
$A l s o T_{2} = T_{1 + 1} = 7_{C_{_{1}}} {(5 x)}^{7 - 1} {(- 2 y)}^{1}$
$= - {7.5}^{6} . {(x)}^{6} .2. y$
   $= - {7.5}^{6} . {(\frac{1}{2})}^{6} .2. \frac{1}{3} (∵ x = \frac{1}{2}, y = \frac{1}{3})$
   $= (\frac{- 14}{3}) {(\frac{5}{2})}^{6}$

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