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# Binomial Expansion watch

1. I'm really not getting all this binomial stuff.

Could someone show me how to solve this?

(i) Expand (2 + x)^7 in ascending powers of x up to and including the term in x³. [4]

(ii) Use your expansion with an appropriate value of x to find an approximate value of 1.99^7. Give you answer to 4 decimal places.

Show your working clearly, giving the numerical value of each term. [3]
Thanks.
2. its in the formula book
:albertein
3. (Original post by Tubbler)
its in the formula book
:albertein
That wasn't much help.
4. formula
(a+b)n =an+nC1 an-1 bx +nC2an-2b2x2 ....etc....bnxn

so that gives you

27 +7c1 26.x +7c2 25.x2 +7c3 24.x3

using a calcualtor for the c values

=

128+448x+672x2 +560x3

will get back to you on part (ii)
5. (Original post by nimajneb)
I'm really not getting all this binomial stuff.

Could someone show me how to solve this?

Thanks.
(2+x)^7 :
(7C0)(2^7)(x°) + (7C1)(2^6)(x^1) + (7C2)(2^5)(x²) + (7C3)(2^4)(x³) + (7C4)(2³)(x^4) + (7C5)(2²)(x^5) + (7C6)(2^1)(x^6) + (7C7)(2^0)(x^7)

where nCr is the binomial coeffecients.

simplifying gives:

128 + 7(64)(x) + 21(32)(x²) + 35(16)(x³) + 35(8)(x^4) + 21(4)(x^5) + 7(2)(x^6) + x^7

further simplification:

128 + 448x + 672x² +560x³ + 280x^4 + 84x^5 + 14x^6 + x^7

This is the full expansion, exams will ask you to state a specific coefficeint of a power of x.

(ii) just substitute the value 1.99 into your expansion, upto the x³ term as you've stated to obtain an estimation. Obviously the more terms of the expansion you use, the more accurate the value.
So if you substitute 1.99 into the full exapnsion of (2+x)^7 above, you'll gain the exact value of (1.99)^7
6. perhaps for part b, take the value of x to be 0 and substitute that in ie the answer will 128, as all the rest become 0.
7. (i) Expand (2 + x)^7 in ascending powers of x up to and including the term in x³. [4]

(ii) Use your expansion with an appropriate value of x to find an approximate value of 1.99^7. Give you answer to 4 decimal places.
Show your working clearly, giving the numerical value of each term. [3]
To use our binomial expansion we want:

8. (Original post by Gaz031)

To use our binomial expansion we want:

wouldnt that give the EXACT value rather than an approx, if she has only figured out up to x3 surely they wouldnt expect the rest to be worked out just to inputa value for x when it is only 3marks, that is why i suggest x as 0 2.00 is close enough to 1.99 and you KNOW all other values are 0 without having to figure out the coeefficients first
9. (Original post by Gaz031)

good work. for the approximation you dont require the dots. it is just the 1st 3 terms which you compute.

wouldnt that give the EXACT value rather than an approx, if she has only figured out up to x3 surely they wouldnt expect the rest to be worked out just to inputa value for x when it is only 3marks, that is why i suggest x as 0 2.00 is close enough to 1.99 and you KNOW all other values are 0 without having to figure out the coeefficients first
no. that would be pointless, since anyone could do 2^7 in their head, without part a.
11. Marisad: The other values aren't 0 but they are very small and can be neglected for the approximation.
Daveman: I wasn't quite sure how to do the symbol in tex(just starting to use it) so I did the equal sign and included the dots (as I didn't want to write something that was clearly false). Thanks for showing me how it's done.

Edit: Thanks!
12. (Original post by chewwy)
no. that would be pointless, since anyone could do 2^7 in their head, without part a.
oh yeah, didn't quite see it like that.
All this brigns back a debate in my maths class, I think the teacher expalined it wrong. dododo *makes excuses for her ignorance* quite pleased i got the first bit right though..maybe p2 won't be quite as big a failure as p3
13. (Original post by Gaz031)
Marisad: The other values aren't 0 but they are very small and can be neglected for the approximation.
Daveman: I wasn't quite sure how to do the symbol in tex(just starting to use it) so I did the equal sign and included the dots (as I didn't want to write something that was clearly false). Thanks for showing me how it's done.

Hmm. The power doesn't seem to be displayed correctly.
Is this how you meant

14. curly brackets not normals ones
15. What am I doing wrong? :s

(i)
(2 + x)^7 = 2^7(1 + ½x)^7
2^7[1 + 7(½x) + 21(½x)² + 35(½x)³]
128 + 448x + 1344x² + 2240x³
16. Anyone?
17. I think you forgot to square and cube the ½ ( in ½x ) when multiplying out.
18. (Original post by Fermat)
I think you forgot to square and cube the ½ ( in ½x ) when multiplying out.
Thanks.

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