In the binomial expansion of (71/2 + 51/3)37, the number of integers are -
Answers
Answer:
Ans : 97
Soln:
From binomial theory we know that the (r+1)th term of the expansion
(a+b)^n is (nCr)*(a^n-r)*(b^r)
In this case a = 5^(1/6) ; b = 2^(1/8) ; n=100 ;
=> Total number of terms = n+1 = 101
For a term to be rational the powers of ‘a’ and ‘b’ should be integral multiples of 6 and 8 respectively so as to cancel out the fractional exponents.
=>n-r = 6k and r = 8m where k and m are some non negative integers
=> r should take values 0,8,16,24,32,40,48,56,64,72,80,88,96 (8m)
=> n-r should take values among 0,6,12,18,24,30,36,42,48,54,60,66,72,78,84,90,96 (6k)
As n= 100 for n-r=6k, r should take 100,94,88,82,76,70,64,58,52,46,40,34,28,22,16,10,4
The common values of r which satisfy r = 8m and n-r = 6k simultaneously are r = 16,40,64,88
=> There will be 4 rational terms
=> The remaining 97 terms will be irrational
Step-by-step explanation:
Answer:
Hello your answer:
Answer
General term of (31/5+71/3)100 is given by,
Tr+1= 100Cr(31/5)100−r(71/3)r
= 100Cr⋅35100−r⋅73r
For a rational term,
5100−r and 3r must be integer.
Hence,
r=0,15,30,45,60,75,90
∴ There are seven rational terms.
Hence, number of irrational terms =101−7=94