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```				   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?

```

7 years ago

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### Answers : (2)

```

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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

```
7 years ago
```										What if the degree is high ? Like here.... Integrals solutions in expansion of (5^1/2 + 7^1/8)^1024 ?
```
one year ago

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