The number of terms in the expansion of (5(1/6) + 7(1/9))1824 which are integers is

Question

The number of terms in the expansion of (5(1/6) + 7(1/9))1824 which are integers is (A) 100 (B) 101 (C) 102 (D) 103

Tardigrade Admin · Accepted Answer

TK + 1=1824CK 5(1824 - K/6) 7(K/9) Where, K=0,1,2.....,1824 Now, (1824 - K/6) will be an integer if K=0,6,12,18,....,1824 Also, (K/9) will be an integer if K=0,9,18,.....,1818. ⇒ Both are integers if K=0,18,36,.....,1818 If the number of terms =n, ⇒ 1818=0+(n - 1)18⇒ n=102

Q. The number of terms in the expansion of $(5^{\frac{1}{6}} + 7^{\frac{1}{9}})^{1824}$ which are integers is

$100$

$101$

$102$

$103$

Q. The number of terms in the expansion of (561​+791​)1824 which are integers is

100

101

102

103

TK+1​=1824CK​561824−K​79K​ Where, K=0,1,2.....,1824 Now, 61824−K​ will be an integer if K=0,6,12,18,....,1824 Also, 9K​ will be an integer if K=0,9,18,.....,1818. ⇒ Both are integers if K=0,18,36,.....,1818 If the number of terms =n, ⇒1818=0+(n−1)18⇒n=102

Q. The number of terms in the expansion of $(5^{\frac{1}{6}} + 7^{\frac{1}{9}})^{1824}$ which are integers is

$100$

$101$

$102$

$103$