Join now for JEE/NEET and also prepare for Boards Learn Science & Maths Concepts for JEE, NEET, CBSE @ Rs. 99! Register Now
Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-5470-145
+91 7353221155
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
OTP to be sent to Change
The number of integral terms in the expansion of (71/2+51/6)78 is
a) 15
b) 14
c) 16
d) 0
I know the ans is b. But I want the method of solving it. What is the condition for finding the no. of integral terms?
Dear Sohini,
This is a good question; a lay man approach has been presented for this problem.
You have,
(71/2 + 51/6 )78
Now, general term of this binomial will be
Tr+1 = 78Cr (71/2)78-r (51/6)r
This general term will be an integer if 78Cr is an integer and 778-r/2 is an integer and 5r/6 is an integer
78Cr will always be a +ve integer
Since 78Cr denotes no. of ways of selecting r things out of 78 things, it cannot be a fraction.
778-r/2 will be an integer
If 78-r/2 is an integer
r = 0 , 2 , 4 , 6 , 8 , 10 , 12 , 14 , 16, ………………..70 , 72 , 74 , 76 and 78 (i)
Also 5 is an integer if r/6 is an integer
r = 0 , 6, 12 , 18 , 24 , ………………… (ii)
Taking the intersection (i) and (ii)
We have,
r = 0 , 6 , 12 , 18 , 24 , 30 , 36 , 42 , 48 , 54, 60 , 66 , 72 , 78
Hence total r permissible is 14
Therefore you have 14 terms
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
Get your questions answered by the expert for free
You will get reply from our expert in sometime.
We will notify you when Our expert answers your question. To View your Question
Win Gift vouchers upto Rs 500/-
Register Yourself for a FREE Demo Class by Top IITians & Medical Experts Today !