The number of integral terms in the expansion of (71/2+51/6)78 isa) 15b) 14c) 16d) 0I 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,

(7^1/2 + 5^1/6 )⁷⁸        

Now, general term of this binomial will be

T_r+1 = ⁷⁸C_r  (7^1/2)^78-r    (5^1/6)^r

This general term will be an integer if ⁷⁸C_r  is an integer and 7^78-r/2 is an integer and 5^r/6 is an integer

⁷⁸C_r will always be a +ve integer

Since ⁷⁸C_r denotes no. of ways of selecting r things out of 78 things, it cannot be a fraction.

7^78-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