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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