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Grade: Upto college level
        

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?


 

9 years ago

Answers : (2)

askIITIians Expert
21 Points
							

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

 

9 years ago
Siddharth Singh
22 Points
							
What if the degree is high ? 
Like here.... Integrals solutions in expansion of (5^1/2 + 7^1/8)^1024 ? 
3 years ago
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