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secondary 3 | A Maths
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I don't understand how to tackle this "Challenge Yourself" question from the Amath 360 TB 3A 2nd Ed.
I'm hoping someone can give me an explanation and enlighting me how to go about deriving and proving the explanation. Thanks!
Date Posted: 3 years ago
Views: 195
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Quite simple. Notice that the 49 has a power of 15 ? And the binomial expansion has a (1 + x) inside. The idea is to break up the 49 into two parts, such that you can perform binomial expansion on it.
49 = 1 + 48 (note that 48 is divisible by 8)
So 49^15 = (1 + 48)^15
= 1^15 + (15choose1)(1^14)(48) + (15choose2)(1^13)(48²) + ... + (15choose13)(1²)(48^13) + (15choose14)(1)(48^14) + 48^15
Every term in the expansion has a multiple of 48 as a factor, except 1^15. So the sum of all those terms except 1^15 are also divisible by 48 (which in turn implies they are divisible by 8 also).
1^15 = 1 so subtracting 1 from 49^15 takes it away. This would make the remaining expression divisible by 8.
49 = 1 + 48 (note that 48 is divisible by 8)
So 49^15 = (1 + 48)^15
= 1^15 + (15choose1)(1^14)(48) + (15choose2)(1^13)(48²) + ... + (15choose13)(1²)(48^13) + (15choose14)(1)(48^14) + 48^15
Every term in the expansion has a multiple of 48 as a factor, except 1^15. So the sum of all those terms except 1^15 are also divisible by 48 (which in turn implies they are divisible by 8 also).
1^15 = 1 so subtracting 1 from 49^15 takes it away. This would make the remaining expression divisible by 8.
Oh it's actually (x + 1)ⁿ so I actually did it the other way round. The sum is the same , just flip the terms in reverse order.
Basically every of the terms, except 1^15, will have a factor of 48 (since 48^r for r >= 1 is divisible by 48).
We must actually also mention that the other terms in the expansion, including nCr, are also integers.
We must actually also mention that the other terms in the expansion, including nCr, are also integers.
Ok thx
There is no need to for the combinations. As for 1^(n-r) , it is self explanatory.
It is understood that for the binomial theorem, the combination (n r) will only be an positive integer since n and r are positive integers and r ≤ n. This is clearly shown in the Pascal's triangle
We do not have to be concerned about the non-integer n and r here (that is under the topic of gamma function)
It is understood that for the binomial theorem, the combination (n r) will only be an positive integer since n and r are positive integers and r ≤ n. This is clearly shown in the Pascal's triangle
We do not have to be concerned about the non-integer n and r here (that is under the topic of gamma function)
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