The Student Room Group

Pascal's Triangle Question

Using Pascal's Triangle only, and not binomial expansion, how would you work this question out:

[ 1 + (x - 1)^2]^3

This is what I have been doing so far, and I know I'm definitely making a mistake because the answer I get is different from the answer in the back of my text book.

My working out is as follows:

[1 + (x - 1)^2]^3

1 + (x - 1)^6

1 + [-1(1 - x)]^6

1 + (1 - x)^6 This part is wrong I think, but I need help here.

1 + (1 + 6(-x) + 15(-x)^2 + 20(-x)^3 + 15(-x)^4 + 6(-x)^5 + 1(-x)^6]

1 + 1 - 6x + 15x^2 - 20x^3 + 15x^4 - 6x^5 + x^6

2 - 6x + 15x^2 - 20x^3 + 15x^4 - 6x^5 + x^6


The actual answer is: 8 - 24x + 36x^2 - 32x^3 + 18x^4 - 6x^5 + x^6


Sorry about the length of this post but I'm getting really :mad: about this!!!
Reply 1
I don't agree with your first step.
Reply 2
Narik
Using Pascal's Triangle only, and not binomial expansion, how would you work this question out:

[ 1 + (x - 1)^2]^3

This is what I have been doing so far, and I know I'm definitely making a mistake because the answer I get is different from the answer in the back of my text book.

My working out is as follows:

[1 + (x - 1)^2]^3

1 + (x - 1)^6

1 + [-1(1 - x)]^6

1 + (1 - x)^6 This part is wrong I think, but I need help here.

1 + (1 + 6(-x) + 15(-x)^2 + 20(-x)^3 + 15(-x)^4 + 6(-x)^5 + 1(-x)^6]

1 + 1 - 6x + 15x^2 - 20x^3 + 15x^4 - 6x^5 + x^6

2 - 6x + 15x^2 - 20x^3 + 15x^4 - 6x^5 + x^6


The actual answer is: 8 - 24x + 36x^2 - 32x^3 + 18x^4 - 6x^5 + x^6


Sorry about the length of this post but I'm getting really :mad: about this!!!

Put it in latex please?
Reply 3
Narik

My working out is as follows:

[1 + (x - 1)^2]^3

1 + (x - 1)^6


(a+b)^3=a^3 + 3 a^2 b + 3 a b^2 + b^3

Do you see the 1 3 3 1 ?

rather than a+b^3 which is what you seem to have done. (or even a^3 + b^3)
Reply 4
Narik
Using Pascal's Triangle only, and not binomial expansion, how would you work this question out:

[ 1 + (x - 1)^2]^3

This is what I have been doing so far, and I know I'm definitely making a mistake because the answer I get is different from the answer in the back of my text book.

Pascals:

...1
..1 1
.1 2 1
1 3 3 1 Thus use this line.
My working out is as follows (AMENDED):

[1+(x1)2]3[br][br][1+x22x+1]3[br][br][x22x+2]3[br][br][x2(2x2)]3[br][br]1(x2)33(x2)2(2x2)1+3(x2)1(2x2)21(2x2)3[br][br]x63x4(2x2)+3x2(4x28x+4)(8x324x2+24x8)[br][br]x66x5+6x4+12x424x3+12x28x3+24x224x+8[br][br]x66x5+18x432x3+36x224x+8[br][br][1 + (x - 1)^2]^3[br][br][1 + x^2-2x+1]^3[br][br][x^2-2x+2]^3[br][br][x^2-(2x-2)]^3[br][br]1(x^2)^3 - 3(x^2)^2(2x-2)^1 + 3(x^2)^1(2x-2)^2 - 1(2x-2)^3[br][br]x^6 - 3x^4(2x-2) + 3x^2(4x^2-8x+4) - (8x^3-24x^2+24x-8)[br][br]x^6 - 6x^5 + 6x^4 + 12x^4 - 24x^3 + 12x^2 - 8x^3 +24x^2 - 24x +8[br][br]x^6 - 6x^5 + 18x^4 -32x^3 + 36x^2 - 24x + 8[br][br]
:smile:
Small123
:smile:
Read the posting guidelines, please. Full solutions tend to be frowned upon. Very nice of you, but not necessarily the most helpful thing to do.

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