Question

Type of terms and corresponding number is written in Column 2 and Column 3 respectively of the Binomial in Column 1.
$\begin{array}{cccccc}& ColumnI& & ColumnII& & Column3\\ \left(I\right)& {(5^{\frac{1}{6}}+7^{\frac{1}{9}})}^{1824}& \left(i\right)& Totalnumberofrationalterms& \left(P\right)& 4\\ \left(II\right)& {(5^{\frac{1}{6}}+2^{\frac{1}{8}})}^{100}& \left(ii\right)& Totalnumberofirrationalterms& \left(Q\right)& 102\\ \left(III\right)& {(3^{\frac{1}{4}}+4^{\frac{1}{3}})}^{99}& \left(iii\right)& \frac{1}{2}\left[\frac{Totalnumberofterms}{Numberofrationalterms}\right],& \left(R\right)& 224\\ & & & where[.]representsgreatest& & \\ & & & integerfunction.& & \\ \left(IV\right)& {(7^{\frac{1}{3}}+{11}^{\frac{1}{9}})}^{2007}& \left(iv\right)& Totalnumberofterms& \left(S\right)& Isdivisibleby13\end{array}$
Select the correct combination

• A. (III) (i) (P)
• B. (IV) (i) (P)
• C. (IV) (iii) (P)
• D. (I) (iv) (R)

Answer 1

The correct option is C (IV) (iii) (P)

(I) $T_{r+1}={}^{1824}{C}_r\left(5^{1/6}{)}^{1824-r}\right(7^{1/9}{)}^r$
$={}^{1824}{C}_r{5}^{304-\frac{r}{6}}{7}^{\frac{r}{9}}$
$\frac{r}{6}$ and $\frac{r}{9}$ must be integers so r must be the LCM of 6 and 9 i.e. 18
r = 0, 18, 36, $\cdots$ 1818
1818 = 0 + ($\lambda$ - 1) 18
102 = Total rational terms
1825 - 102 = 1723 = total irrational terms
Total number of terms = 1825

(II)$T_{r+1}={}^{100}{C}_r(5^{1/6}{)}^{100-r}{2}^{\frac{r}{8}}$
$\frac{100-r}{6}$ and $\frac{r}{8}$ must be integers for rational terms
r = 0, 8, 16, 24 $\cdots$ 88, 96
100 - r = 100, 92, 84 $\cdots$ 12, 4
r = 16, 40, 64, 88 for $\frac{100-r}{6}$ and $\frac{r}{8}$ to be integers
So, total rational terms = 4
total terms = 101
total irrational terms = 97

(III) ${(3^{\frac{1}{4}}+4^{\frac{1}{3}})}^{99}={(3^{\frac{1}{4}}+2^{\frac{2}{3}})}^{99}$
$T_{r+1}={}^{99}{C}_r{\left(3^{\frac{1}{4}}\right)}^{99-r}{2}^{\frac{2r}{3}}$
for rational terms
$\frac{99-r}{4}$ and $\frac{2r}{3}$ must be integers
r = 0, 3, 6 $\cdots$ 99
99 - r = 99, 96, 93 $\cdots$ 3, 0
$\frac{99-r}{4}$ and $\frac{2r}{3}$ must be integers for rational terms
Hence 99 - r = 0, 12, 24, 36, 48, 60, 72, 84, 96
Total rational terms = 9
Total terms = 100
Hence irrational terms = 91

(IV) $T_{r+1}={}^{2007}{C}_r{7}^{\frac{2007-r}{3}}{11}^{\frac{r}{9}}$;for rational terms $\frac{2007-r}{3}$, $\frac{r}{9}$ must be integers
r = 0, 9, 18, $\cdots$ 2007
So, total rational terms = 224
Total irrational terms = 2008 - 224 = 1784
Total terms = 2008