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Binomial Expansion

I'm really not getting all this binomial stuff. :frown:

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.

Reply 1

Tubbler

its in the formula book
:albertein

Reply 2

Ben

OP

12

Tubbler

its in the formula book
:albertein

That wasn't much help. :confused:

Reply 3

Frogsofhope

2

formula
(a+b)ⁿ =aⁿ+ⁿC₁ a^n-1 bx +ⁿC₂a^n-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)

Reply 4

Vijay1

nimajneb

I'm really not getting all this binomial stuff. :frown:

Could someone show me how to solve this?

Thanks. :smile:

(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

Reply 5

Frogsofhope

2

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.

Reply 6

Gaz031

15

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

Unparseable latex formula:

(2+x)^7\\[br]=(2^7)(1+\frac{x}{2})^7\\[br]=(128)[1+7(\frac{x}{2})+21(\frac{x}{2})^2+35(\frac{x}{2})^3....]\\[br]=128+448x+672x^2+560x^3...\\[br]

(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:

Unparseable latex formula:

2+x=1.99\\[br]x=-0.01\\[br]1.99^7=128+448(-.01)+672(-.01)^2+560(-.01)^3...

Evaluate with your calculator.

Reply 7

Frogsofhope

2

Gaz031

Unparseable latex formula:

(2+x)^7\\[br]=(2^7)(1+\frac{x}{2})^7\\[br]=(128)[1+7(\frac{x}{2})+21(\frac{x}{2})^2+35(\frac{x}{2})^3....]\\[br]=128+448x+672x^2+560x^3...\\[br]

To use our binomial expansion we want:

Unparseable latex formula:

2+x=1.99\\[br]x=-0.01\\[br]1.99^7=128+448(-.01)+672(-.01)^2+560(-.01)^3...

Evaluate with your calculator.

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

Reply 8

DaveManUK

10

Gaz031

Unparseable latex formula:

2+x=1.99\\[br]x=-0.01\\[br]1.99^7=128+448(-.01)+672(-.01)^2+560(-.01)^3[b]...[/b]

Evaluate with your calculator.

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

1.99^7\approx128+448(-.01)+672(-.01)^2+560(-.01)^3

Reply 9

Chewwy

17

marisad_uk

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.

Reply 10

Gaz031

15

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

\approx

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.

Unparseable latex formula:

tanx \approx x\\[br][br](1+x)^{\frac{1}{2}} \approx 1+\frac{1}{2}x-\frac{1}{8}x^2

Edit: Thanks!

Reply 11

Frogsofhope

2

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. :banghead:

All this brigns back a debate in my maths class, I think the teacher expalined it wrong. dododo *makes excuses for her ignorance* :biggrin:

quite pleased i got the first bit right though..maybe p2 won't be quite as big a failure as p3

Reply 12

Fermat

8

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

\approx

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.

Unparseable latex formula:

tanx \approx x\\[br][br](1+x)^(\frac{1}{2}) \approx 1+\frac{1}{2}x-\frac{1}{8}x^2

Hmm. The power doesn't seem to be displayed correctly.