Question

Number of Irrational terms in the Expansion of
(2^1/3+3^1/2+5^1/6)^10
is equal to :-
(A) 3
(B) 60
(C)63
(D) 66​

Answer 1

Given:

(2^1/3+3^1/2+5^1/6)^10

To Find:

Number of Irrational terms in the Expansion of  (2^1/3+3^1/2+5^1/6)^10

Solution:

Binomial expansion for three terms ,

• ( a + b + c )^n =  ∑ $\frac{n!}{i! j! k!}$ $a^{i} b^{j} c^{k}$  , i + j +  k = 0

Applying this in the expansion of 2^1/3+3^1/2+5^1/6)^10 ,

• We can find the number of rational terms.
• Subtract it from total number of terms.
• For rational terms.
• i  should be a multiple of 3 = 0 ,3 ,6, 9
• j should be a multiple of 2 = 0 , 2, 4, 6 ,8 ,10
• k should be a multiple of 6 = 0 ,6 .
• Also i , j , k should add to 10.
• Possible solutions: for ( i , j , k )
• ( 0 , 10 , 0 ) , ( 0 , 4 , 6 ) , ( 6 , 4 , 0 )  = 3 terms.

These 3 terms are:

• ( 0 , 10 , 0 ) :  $2^{0}3^{5} 5^{0}$

• ( 0 , 4 , 6 ) : $2^{0} 3^{2} 5^{1}$

• ( 6 , 4 , 0 )  : $2^{2} 3^{2} 5^{0}$

Total number of terms in binomial expansion of three terms  = (n+1)(n+2)/2

• Total number of terms = 11 x 12 /2 = 66
• Total number of irrational terms = 66 - 3 == 66.

Number of Irrational terms in the Expansion of  (2^1/3+3^1/2+5^1/6)^10  is equal to :- 63