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pls help! the topic is binomial expansion.
Date Posted: 3 years ago
Views: 252
The trick is to rewrite one of the terms such that it is a sum/difference of 3 and another number.
Then expand using binomial theorem.
5^39 - 2^39
= (2 + 3)^39 - 2^39
= 2^39 + (39choose1)(2^38)(3¹) + (39choose2)(2^37)(3²) + ... + (39choose37)(2¹)(3^38) + 3^39 - 2^39
= (39choose1)(2^38)(3¹) + (39choose2)(2^37)(3²) + ... + (39choose37)(2¹)(3^38) + 3^39
= 3 [(39choose1)(2^37) + (39choose2)(2^38)(3) + ... + (39choose37)(2¹)(3^37) + 3^38]
Notice that the two 2^39 cancelled out (since 2^39 - 2^39 = 0)
And each of the remaining terms has factor that is a multiple of 3. We can factor 3 out.
The expression is now a multiple of 3 since all the terms in the brackets are all positive integers
(their factors are multiples of 2 or 3, the combinations are all positive integers as seen in the Pascal's triangle)
So the expression is divisible by 3.
Then expand using binomial theorem.
5^39 - 2^39
= (2 + 3)^39 - 2^39
= 2^39 + (39choose1)(2^38)(3¹) + (39choose2)(2^37)(3²) + ... + (39choose37)(2¹)(3^38) + 3^39 - 2^39
= (39choose1)(2^38)(3¹) + (39choose2)(2^37)(3²) + ... + (39choose37)(2¹)(3^38) + 3^39
= 3 [(39choose1)(2^37) + (39choose2)(2^38)(3) + ... + (39choose37)(2¹)(3^37) + 3^38]
Notice that the two 2^39 cancelled out (since 2^39 - 2^39 = 0)
And each of the remaining terms has factor that is a multiple of 3. We can factor 3 out.
The expression is now a multiple of 3 since all the terms in the brackets are all positive integers
(their factors are multiples of 2 or 3, the combinations are all positive integers as seen in the Pascal's triangle)
So the expression is divisible by 3.
b)
Similarly,
2^99 + 3^99
= (5 - 3)^99 + 3^99
= 5^99 + (99choose1)(5^98)(-3)¹ + (99choose2)(5^97)(-3)² + ... + (99choose98)(5¹)(-3)^98 + (-3)^99 + 3^99
= 5^99 + (99choose1)(5^98)(-3) + (99choose2)(5^97)(9) + ... + (99choose98)(5¹)(3)^98 - 3^99 + 3^99
(Because (-3)^99
= (3 x -1)^99
= 3^99 x (-1)^99
= 3^99 x -1, since -1 to the power of an odd integer is just itself.
Notice that (-1)^99 = -1 x -1 x -1 x -1 x ... -1 for a total of 99 "-1". And every pair's negative signs cancel out except the last, which is unpaired)
= 5^99 + (99choose1)(5^98)(-3) + (99choose2)(5^97)(9) + ... + (99choose98)(5¹)(3)^98
= 5[5^98 + (99choose1)(5^97)(-3) + (99choose2)(5^96)(9) + ... + (99choose98)](3)^98
5 can also be factored out.
Similarly,
2^99 + 3^99
= (5 - 3)^99 + 3^99
= 5^99 + (99choose1)(5^98)(-3)¹ + (99choose2)(5^97)(-3)² + ... + (99choose98)(5¹)(-3)^98 + (-3)^99 + 3^99
= 5^99 + (99choose1)(5^98)(-3) + (99choose2)(5^97)(9) + ... + (99choose98)(5¹)(3)^98 - 3^99 + 3^99
(Because (-3)^99
= (3 x -1)^99
= 3^99 x (-1)^99
= 3^99 x -1, since -1 to the power of an odd integer is just itself.
Notice that (-1)^99 = -1 x -1 x -1 x -1 x ... -1 for a total of 99 "-1". And every pair's negative signs cancel out except the last, which is unpaired)
= 5^99 + (99choose1)(5^98)(-3) + (99choose2)(5^97)(9) + ... + (99choose98)(5¹)(3)^98
= 5[5^98 + (99choose1)(5^97)(-3) + (99choose2)(5^96)(9) + ... + (99choose98)](3)^98
5 can also be factored out.
Alternative to a) using remainder theorem/factor theorem :
Consider the polynomial 5^39 - x^39
Let f(x) = 5^39 - x^39
When f(x) is divided by (x - 5), the remainder is f(5)
= 5^39 - 5^39
= 0
Since f(5) = 0, by the Factor Theorem, (x - 5) is a factor of f(x)
So f(x) = Q(x) (x - 5), where (Q(x) is a polynomial and the quotient.
Since x - 5 is always an integer for integer values of x, Q(x) will also be an integer since (5^39 - x^39) would be an integer for these values of x, and the product of two integers gives another integer)
Sub x = 2,
5^39 - 2^39 = Q(2) (2 - 5)
= -3 Q(2)
= 3 (-Q(2))
(5^39 is bigger than 2^39 so 5^39 - 2^39 >0 . This means that Q(2) is negative and -Q(2) is positive)
We have factored out the 3 and so it is divisible by 3 since -Q(2) is a positive integer.
Consider the polynomial 5^39 - x^39
Let f(x) = 5^39 - x^39
When f(x) is divided by (x - 5), the remainder is f(5)
= 5^39 - 5^39
= 0
Since f(5) = 0, by the Factor Theorem, (x - 5) is a factor of f(x)
So f(x) = Q(x) (x - 5), where (Q(x) is a polynomial and the quotient.
Since x - 5 is always an integer for integer values of x, Q(x) will also be an integer since (5^39 - x^39) would be an integer for these values of x, and the product of two integers gives another integer)
Sub x = 2,
5^39 - 2^39 = Q(2) (2 - 5)
= -3 Q(2)
= 3 (-Q(2))
(5^39 is bigger than 2^39 so 5^39 - 2^39 >0 . This means that Q(2) is negative and -Q(2) is positive)
We have factored out the 3 and so it is divisible by 3 since -Q(2) is a positive integer.
thanks for the explanation, it’s very helpful and detailed! but can you please tell me which formula you used to expand each equation?
Welcome. I have used the formula for the binomial theorem , which is found in your textbooks and also the formula sheet for O level Additional Maths
If you want an online version it can be found at the SEAB website :
https://www.seab.gov.sg/docs/default-source/national-examinations/syllabus/olevel/2021syllabus/4049_y21_sy.pdf
https://www.seab.gov.sg/docs/default-source/national-examinations/syllabus/olevel/2021syllabus/4049_y21_sy.pdf
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there are definitely easier ways to show the requested result, but since the topic is on binomial expansion, then just have to use binomial expansion.
Date Posted:
3 years ago
in general, to show that a is divisible by b , just have to show that a = b x c, where c is an integer.
thanks for the answer, it’s very helpful snd easy to understand! but could you please tell me which formula you used to expand the equation? thanks again <3
Am using the standard binomial expansion which is …
(a+b)^n
= a^n
+ [n 1] x (a)^(n-1) x (b)
+ [n 2] x (a)^(n-2) x (b)^2
+ [n 3] x (a)^(n-3) x (b)^3
+ …..
+ [n r] x (a)^(n-r) x (b)^r
+ ….
+ b^n
where [n r] = n!/[r!(n-r)!]
sorry, a bit difficult to type formulas properly in these comments. for [n r] you should write the n above the r, but not able to type that here.
(a+b)^n
= a^n
+ [n 1] x (a)^(n-1) x (b)
+ [n 2] x (a)^(n-2) x (b)^2
+ [n 3] x (a)^(n-3) x (b)^3
+ …..
+ [n r] x (a)^(n-r) x (b)^r
+ ….
+ b^n
where [n r] = n!/[r!(n-r)!]
sorry, a bit difficult to type formulas properly in these comments. for [n r] you should write the n above the r, but not able to type that here.
alternative way to show the results, without using binomial expansion, would be to consider remainders.
for (a),
when you divide 5,25,125,625,3125,... (powers of 5) ...
by 3, you get remainder 2,1,2,1,2,... (repeating pattern 2,1)
when you divide 2,4,8,16,32,... (powers of 2) ...
by 3, you get remainder 2,1,2,1,2,... (repeating pattern 2,1)
so comparing 5^n and 2^n, when divided by 3, the remainders are always same regardless of value of n.
since the remainders cancel out each other,
so actually (5^n-2^n) is divisible by 3, for all n.
for (b),
when you divide 2,4,8,16,32, … (powers of 2) … by 5, you get 2,4,3,1,2,4,3,1,... (repeating pattern 2,4,3,1)
when you divide 3,9,27,81,243, … (powers of 3) … by 5, you get 3,4,2,1,3,4,2,1,... (repeating pattern 3,4,2,1)
since 99 divide by 4 is remainder 3, so …
2^99 divide by 5 will have remainder 3
3^99 divide by 5 will have remainder 2.
since the remainders add up to 5, this means that (2^99 + 3^99) is divisible by 5.
the rest (c)(d)(e) can be shown the same way.
for (a),
when you divide 5,25,125,625,3125,... (powers of 5) ...
by 3, you get remainder 2,1,2,1,2,... (repeating pattern 2,1)
when you divide 2,4,8,16,32,... (powers of 2) ...
by 3, you get remainder 2,1,2,1,2,... (repeating pattern 2,1)
so comparing 5^n and 2^n, when divided by 3, the remainders are always same regardless of value of n.
since the remainders cancel out each other,
so actually (5^n-2^n) is divisible by 3, for all n.
for (b),
when you divide 2,4,8,16,32, … (powers of 2) … by 5, you get 2,4,3,1,2,4,3,1,... (repeating pattern 2,4,3,1)
when you divide 3,9,27,81,243, … (powers of 3) … by 5, you get 3,4,2,1,3,4,2,1,... (repeating pattern 3,4,2,1)
since 99 divide by 4 is remainder 3, so …
2^99 divide by 5 will have remainder 3
3^99 divide by 5 will have remainder 2.
since the remainders add up to 5, this means that (2^99 + 3^99) is divisible by 5.
the rest (c)(d)(e) can be shown the same way.
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