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7.Binomial Theorem
medium
${\left( {\sqrt 3 + \sqrt[8]{5}} \right)^{256}}$ ના વિસ્તરણમાં પૂર્ણાક પદની સંખ્યા મેળવો.
A
$32$
B
$33$
C
$34$
D
$35$
(AIEEE-2003)
Solution
(b) ${T_{r + 1}} = {}^{256}{C_r}{(\sqrt 3 )^{256 – r}}{(\sqrt[8]{5})^r}$ $ = {}^{256}{C_r}{(3)^{\frac{{256 – r}}{2}}}{(5)^{r/8}}$
Terms would be integral if $\frac{{256 – r}}{2}$ and $\frac{r}{8}$ both are positive integer.
As $0 \leq r \leq 256$, $\therefore r = 0,\,8,\,16,\,24,…..,256$
For above values of $r$, $\left( {\frac{{256 – r}}{2}} \right)$ is also an integer.
Total number of values of $r = 33.$
Standard 11
Mathematics
