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All You Need To Know About Binomial Expansion 2/2

4 Types of Binomial Theorem Questions

‘O’ Level A Maths binomial questions (For more info on the basic formulas to memorize and understand, click here)

  1. Expand and Multiply
  2. Expand and Substitute
  3. Finding Specific Terms
  4. When n is Unknown …
  5. Summary Exercise

[−] A. Expand and Multiply

This is a basic binomial question where you expand an expression up to the first few terms. You may or may not be told how many terms to expand to, but if the powers are high, ask yourself if there is a need to expand everything.

Binomial Expansion Question Type 1:

Find, in ascending powers of x, the first 3 terms in the expansion of
(a) (1 + 4x)6 (b) (1 − 2x)14

Hence find the expansion of (1 + 4x)6(1 − 2x)14 up to the terms in x2.

ANS: Using the General Binomial Equation given in the formula sheet,

(a) (1 + 4x)6
= 16 + 6C1(15)(4x) + 6C2(14)(4x)2 + …
= 1 + 6(1)(4x) + 15(1)(16x2) + …
= 1 + 24x + 240x2 + …

(b) (1 − 2x)14
= 114 + 14C1(113)(−2x) + 14C2(112)(−2x)2 + …
= 1 − 14(1)(2x) + 91(1)(4x2) + …
= 1 − 28x + 364x2 + …

To obtain the first 3 terms of (1 + 4x)6(1 − 2x)14, we multiply and expand (1 + 24x + 240x2 + …)(1 − 28x + 364x2 + …)

Since we are only interested in terms up to x2, we ignore all combos that result in powers of x more than 2 so as to save time!

(1 + 24x + 240x2 + …)(1 − 28x + 364x2 + …)
= 1(1 − 28x + 364x2) + 24x(1 − 28x + 364x2) + 240x2(1 − 28x + 364x2)
= 1 − 28x + 364x2 + 24x − 672x2 + 240x2
1 − 4x − 68x2

Sometimes the question don’t specify the number of terms to expand to, e.g. “expand up to and including the terms in x3” or “given that (1 + 4x)6(1 − 2x)14 = a+ bx + cx2 + …, find a, b and c.”

In such cases, if the powers are high, ask yourself if there is a need to expand everything.


Try and See if you’v understood Revision Exercise

(2+1/{2x})^2(1-2x)^7

Find the term independent of x in the expansion .

[Answer: −3]


(x^3 - 2/{x^2})^10

This is an example of a more challenging binomial question since it consists of more than one x and inverse powers of x e.g. 


[−] B. Expand and Substitute

Binomial Expansion Question Type 2:

Expand (1 − p)5. Use this result to find the expansion of (1 − x + x2)5 as far as the term in x3. Use a suitable value of x to estimate (1.11)5 from this expansion.

ANS: Using the General Binomial Equation given in the formula sheet,

(1 − p)5
= 15 + 5C1(14)(−p) + 5C2(13)(−p)2 + 5C3(12)(−p)3 + 5C4(12)(−p)4 + (−p)5
= 1 − 5p + 10p2 − 10p3 + 5p4 − p5

Since (1 − x + x2)5 is similar to (1 − p)5 = 1 − 5p + 10p2 + …

⇒ 1 − p = 1 − x + x2
⇒ p = x − x2 → our substitute expression!

Substitute p = x − x2 into the expansion:

= 1 − 5(x − x2) + 10(x − x2)2 − 10(x − x2)3 + 5(x − x2)4 − (x − x2)5

Since we are only interested in terms up to x3, we ignore all combos that result in powers of x more than 3 to save time

= 1 − 5x + 5x2 + 10(x2 − 2x3 + x4) − 10(x3 + …)

There is no need to expand that cubic expression as there’s only one x3 term in (x2 − 2x3 + x4)(x − x2)

= 1 − 5x + 5x2 + 10x2 − 20x3 − 10x3 + …
= 1 − 5x + 15x2 − 30x3 + …

To estimate (1.11)5 using x, we equate 1.11 with 1 − x + x2

⇒ 1.11 = 1 − x + x2
⇒ x2 − x − 0.11 = 0
Solving the quadratic equation, x = −0.1 or 1.1

In order to use (1 − x + x2)5 = 1 − 5x + 15x2 − 30x3 + … to estimate (1.11)5, it’s better to use x = −0.1 since its higher powers are comparatively smaller enough to be ignored i.e. −0.14 = −0.0001 vs 1.14 = 1.4641

∴ (1.11)5 ≈ 1 − 5(−0.1) + 15(−0.1)2 − 30(−0.1)3 ≈ 1.68


[−] C. Finding Specific Terms

Binomial Expansion Question Type 3:

This is a straightforward (a + b)n question, don’t expand everything …

ANS:
Using the Tr+1 equation
T_{r+1} = (matrix{2}{1}{10 r})(x^3)^{10-r}(-2/{x^2})^r
{} = (matrix{2}{1}{10 r})x^{30-3r} (-2)^r x^{-2r}
{} = (matrix{2}{1}{10 r}) (-2)^r x^{30-5r}

The key is to first obtain the Tr+1 ‘formula’ and then arrange and separately nicely the constants from the x with the help of indices :

⇒ power of x = 30 − 5r

(a) For x10, 30 − 5r = 10 ⇒ r = 4

T_5 = (matrix{2}{1}{10 4})(-2)^4 x^10 = 3360x^10

Term in x10 = 3360 x10

(b) For x−5, 30 − 5r = −5 ⇒ r = 7

(matrix{2}{1}{10 7})(-2)^7 = -15360

Coefficient of x−5 = -15360

(c) For the constant term x0 , 30 − 5r = 0 ⇒ r = 6

T_7 = (matrix{2}{1}{10 6})(-2)^6 x^0 = 13 440

Constant Term = 13440


Try and See if you’v understood Revision Exercise

[−] D. When n is Unknown …

Whenever n is unknown, it’s almost certain that you’ll have to expand the nCr Formula

{n(n-1) cdots (n-r+1)}/{r!}

into the  form within an expansion or Tr+1 formula (that nCrcalculator button will suddenly become useless for those who like to keep pressing it), to obtain the value of n and some other variables via the comparison of the coefficients of suitable terms.

Binomial Expansion Question Type 4:

Given that (1 + ax)n = 1 − 12x + 63x2 + …, find a and n.

ANS: (1 + ax)n = 1n + nC11n−1(ax) + nC21n−2(ax)2 + …

we only have to expand until first 3 terms to include

1) a constant

2) x

3) x2 

Since we have to solve for n, it will be best to convert to the above nCr formula.

Compare coefficients of x

na = −12

{n(n-1)}/{2!}a^2 = 63

Compare coefficients of x2

{n(n-1)}/{2}(-12/n)^2 = 63

Sub a = −12/n

9n = 72
∴ n = 8
Sub n = 72 into (1):
∴ a = −12/8 = 3/2

Try and See if you’v understood Revision Exercise

In the expansion of (2 + 3x)n, the coefficients of x3 and x4 are in the ratio 8 : 15. Find the value of n.

[Answer: n = 8]

[−] Summary Exercise

This question from the 2009 GCE ‘O’ Level AMaths Paper 2 had resulted in crazy hair pulling among those who took the paper that year. You should be able to solve it easily now 🙂 If you don’t know how to do this question, click here 🙁

(i) Write down the first three terms in the expansion, in ascending powers of x, of 

(2-x/4)^n

, where n is a positive integer greater than 2.

The first two terms in the expansion, in ascending powers of x, of  

(1+x)(2-x/4)^n

are a + bx2, where a and b are constants.

(ii) Find the value of n.
(iii) Hence find the value of a and b.

[Answer: n = 8, a = 256, b = −144]

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