7.Binomial Theorem
hard

જો ${\left( {\sqrt[3]{{\frac{a}{{\sqrt b }}}} + \sqrt {\frac{b}{{\sqrt[3]{a}}}} } \right)^{21}}$ ના વિસ્તરણમાં ${(r + 1)^{th}}$ ના પદમાં $a$ અને $b$ ની ઘાતાંક સમાન હોય , તો $r$ મેળવો.

A

$9$

B

$10$

C

$8$

D

$6$

Solution

(a) We have ${T_{r + 1}} = {\,^{21}}{C_r}{\left( {\sqrt[3]{{\frac{a}{{\sqrt b }}}}} \right)^{21 – r}}{\left( {\sqrt {\frac{b}{{\sqrt[3]{a}}}} } \right)^r}$

$ = {\,^{21}}{C_r}{a^{7 – (r/2)}}{b^{(2/3)r – (7/2)}}$

Since the powers of $a$ and $b$ are the same,

$\therefore$ $7 – \frac{r}{2} = \frac{2}{3}r – \frac{7}{2}$

$\Rightarrow r = 9$

Standard 11
Mathematics

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