Chapter 3: Sequences and Series
Learning Outcomes
(a) Find the expansion of where is a positive integer.
(b) Write notations and as a binomial coefficient.
(c) Determine the general term in a binomial expansion where is a positive
integer. Bloom: Understanding 𝒂 + 𝒃 𝒏 𝒏 𝒏! 𝒏 𝒓 𝒂 + 𝒃 𝒏 𝒏 𝑛𝐶 𝑟 =
Learning Outcomes
(d) Determine the expansion of for
where is a rational number for both positive and negative numbers.
* where
* Use binomial expansion to approximate values such as . Bloom: Understanding 𝟏 + 𝒙 𝒏 𝒙 < 𝟏 𝒏 𝒂 + 𝒃 𝒏 = 𝒂𝒏 𝟏 + 𝒂 𝒃 𝒏 𝒂 > 𝒃 𝟐, 𝟏. 𝟎𝟎𝟏 𝟏𝟎, 𝟑 𝟓
Binomial Expansion
Bloom: Remembering 𝒏! = 𝒏 𝒏 − 𝟏 𝒏 − 𝟐 … 𝟑 × 𝟐 × 𝟏 𝒏 𝒓 = 𝒏𝑪 𝒓 = 𝒏! 𝒓! 𝒏 − 𝒓 !If is a positive integer, the general result for the expansion of is 𝒏 𝒂 + 𝒃 𝒏 𝒂 + 𝒃 𝒏 = 𝒏 𝟎 𝒂 𝒏 + 𝒏 𝟏 𝒂 𝒏−𝟏𝒃 + 𝒏 𝟐 𝒂 𝒏−𝟐𝒃𝟐 + 𝒏 𝟑 𝒂 𝒏−𝟑𝒃𝟑 + … 𝒏 𝒓 𝒂 𝒏−𝒓𝒃𝒓 + ⋯ + 𝒏 𝒏 𝒃 𝒏 Type 1
Binomial Expansion
Bloom: Remembering
kwkan@KMK
If is negative or fractional, the binomial series for is given by 𝒏 𝟏 + 𝒙 𝒏 𝟏 + 𝒙 𝒏 = 𝟏 + 𝒏𝒙 + 𝒏(𝒏 − 𝟏)𝒙 𝟐 𝟐! + 𝒏(𝒏 − 𝟏)(𝒏 − 𝟐)𝒙𝟑 𝟑! + … Type 2 𝑻𝒓+𝟏 = 𝒏𝑪𝒓𝒂𝒏−𝒓𝒃𝒓
The general term,
𝒂 + 𝒃𝒙 𝒏 = 𝒂𝒏 𝟏 + 𝒃 𝒂
𝒏
before applying the expansion of as shown above.
𝟏 + 𝒙 𝒏
Example
1. Expand the following using the binomial theorem.
(a) (b)
2. Find the 12th term in the expansion of
as a series in ascending powers of .
Bloom: Understanding
𝒙 + 𝒚 𝟑 𝒙 − 𝒚 𝟒
𝟏 + 𝒙 𝟐𝟎 𝒙
Example
3. Find the coefficient of in the expansion of .
4. Find the term independent of in the expansion of . Bloom: Understanding 𝒙𝟔 𝟑 + 𝒙 𝟏𝟐 𝒙 𝒙 − 𝟏 𝒙 𝟏𝟐
Solution
1. (a) (b) Bloom: Understanding 𝒙 + 𝒚 𝟑 = 𝒙𝟑 + 𝟑 𝟏 𝒙 𝟐 𝒚 + 𝟑 𝟐 𝒙 𝒚 𝟐 + 𝒚 𝟑 = 𝒙𝟑 + 𝟑𝒙𝟐𝒚 + 𝟑𝒙𝒚𝟐 + 𝒚𝟑 𝒙 − 𝒚 𝟒 = 𝒙𝟒 + 𝟒 𝟏 𝒙 𝟑 −𝒚 + 𝟒 𝟐 𝒙 𝟐 −𝒚𝟐 + 𝟒 𝟑 𝒙 −𝒚 𝟑 + −𝒚 𝟒 = 𝒙𝟒 − 𝟒𝒙𝟑𝒚 + 𝟔𝒙𝟐𝒚𝟐 − 𝟒𝒙𝒚𝟑 + 𝒚𝟒Solution (continue…)
2.
3.
Bloom: Understanding
𝑻𝒓+𝟏 = 𝒏𝑪𝒓𝒂𝒏−𝒓𝒃𝒓 Using 𝑻𝒓+𝟏 term formula 𝑻𝟏𝟐 = 𝟐𝟎 𝟏𝟏 𝟏 𝟐𝟎−𝟏𝟏 𝒙 𝟏𝟏 = 𝟏𝟔𝟕𝟗𝟔𝟎𝒙𝟏𝟏 𝑻𝒓+𝟏 = 𝟏𝟐 𝒓 𝟑 𝟏𝟐−𝒓 𝒙 𝒓 Using 𝑻 𝒓+𝟏 term formula 𝑻𝟕 = 𝟏𝟐 𝟔 𝟑 𝟏𝟐−𝟔 𝒙 𝟔 Obviously, r=6 = 𝟔𝟕𝟑𝟓𝟗𝟔𝒙𝟔 The coeficient of is . 𝒙𝟔 𝟔𝟕𝟑𝟓𝟗𝟔
Solution (continue…)
4.
Bloom: Understanding Independent of means without term. This is only
make possible if the power of is zero.
𝑻𝒓+𝟏 = 𝟏𝟐 𝒓 𝒙 𝟏𝟐−𝒓 −𝟏 𝒙 𝒓 𝒙 𝒙 𝒙
Collect all terms in and simplifying them if possible.𝒙
𝒙𝟏𝟐−𝒓 𝒙𝒓 = 𝒙 𝟏𝟐−𝟐𝒓 = 𝒙𝟎 ∴ 𝟏𝟐 − 𝟐𝒓 = 𝟎 𝒓 = 𝟔 𝑺𝒐, 𝑻𝟕 = 𝟏𝟐 𝟔 𝒙 𝟏𝟐−𝟔 − 𝟏 𝒙 𝟔 = 𝟗𝟐𝟒
Self-check
Bloom: Applying
1. Expand the following using the binomial theorem.
(a) (b)
2. Find the 10th term in the expansion of
as a series in ascending powers of .
𝒙 + 𝒚 𝟒 𝟏 − 𝒙 𝟒
𝟏 + 𝟐𝒙 𝟏𝟖 𝒙
Self-check
Bloom: Applying
3. Find the coefficient of in the expansion of .
4. Find the term independent of in the expansion of . 𝒙𝟔 𝟐 + 𝒙 𝟐 𝟏𝟎 𝒙 𝒙 + 𝟏 𝟐𝒙𝟐 𝟏𝟐
Answer Self-check
(1) (a) (b) (2) (3) 52.5 (4) Bloom: Applying 𝒙𝟒 + 𝟒𝒙𝟑𝒚 + 𝟔𝒙𝟐𝒚𝟐 + 𝟒𝒙𝒚𝟑 + 𝒚𝟒 𝟏 − 𝟒𝒙 + 𝟔𝒙𝟐 − 𝟒𝒙𝟑 + 𝒙𝟒 𝑻𝟏𝟎 = 𝟐𝟒𝟖𝟗𝟑𝟒𝟒𝟎𝒙𝟗 𝟒𝟗𝟓 𝟏𝟔Example
1. Expand the following up to the term in . (a)
(b)
2. For which value of are the expansions in Question 1 valid? Bloom: Understanding 𝟏 + 𝒙 𝟏𝟐 𝒙𝟑 𝟏 − 𝒙 𝟐 𝟏 𝟒 𝒙
Example
3. Expand the function in a series of
ascending powers of as far as the term in . State the value of for which each
expansion is valid.
4. Obtain the expansion of up to the term. By putting , find a rational
approximation for . Bloom: Understanding 𝟗 + 𝒙 𝟏𝟐 𝒙𝟑 𝒙 𝒙 𝟏 − 𝟐𝒙 𝒙𝟐 𝒙 = 𝟏 𝟏𝟎𝟎 𝟐
Solution
1. (a) Bloom: Understanding 𝟏 + 𝒙 𝟏𝟐 = 𝟏 + 𝟏 𝟐 𝒙 + 𝟏 𝟐 𝟏 𝟐 − 𝟏 𝟐! 𝒙 𝟐 + 𝟏 𝟐 𝟏 𝟐 − 𝟏 𝟏 𝟐 − 𝟐 𝟑! 𝒙 𝟑 + … = 𝟏 + 𝟏 𝟐 𝒙 − 𝟏 𝟖 𝒙 𝟐 + 𝟏 𝟏𝟔 𝒙 𝟑 + …Solution (Continue…)
1. (b) Bloom: Understanding + 𝟏 𝟒 𝟏 𝟒 − 𝟏 𝟏 𝟒 − 𝟐 𝟑! − 𝒙 𝟐 𝟑 + … = 𝟏 − 𝟏 𝟖 𝒙 − 𝟑 𝟏𝟐𝟖 𝒙 𝟐 − 𝟕 𝟏𝟎𝟐𝟒 𝒙 𝟑 + … 𝟏 − 𝒙 𝟐 𝟏 𝟒 = 𝟏 + 𝟏 𝟒 − 𝒙 𝟐 + 𝟏 𝟒 𝟏 𝟒 − 𝟏 𝟐! − 𝒙 𝟐 𝟐Solution (Continue…)
2. (a) For expansion is valid,
(b) For expansion is valid,
Bloom: Understanding 𝒙 < 𝟏 ∴ −𝟏 < 𝒙 < 𝟏 𝒙 𝟐 < 𝟏 ∴ −𝟐 < 𝒙 < 𝟐 −𝟏 < 𝒙 𝟐 < 𝟏
Solution (Continue…)
3. Bloom: Understanding 𝟗 + 𝒙 𝟏𝟐 = 𝟑 𝟏 + 𝒙 𝟗 𝟏 𝟐 = 𝟑 𝟏 + 𝟏 𝟐 𝒙 𝟗 + 𝟏 𝟐 𝟏 𝟐−𝟏 𝟐! 𝒙 𝟗 𝟐 + 𝟏 𝟐 𝟏 𝟐−𝟏 𝟏 𝟐−𝟐 𝟑! 𝒙 𝟗 𝟑 + … = 𝟗 𝟏 + 𝒙 𝟗 𝟏 𝟐 = 𝟑 + 𝒙 𝟔 − 𝟏 𝟐𝟏𝟔 𝒙 𝟐 + 𝒙 𝟑 𝟑𝟖𝟖𝟖 + ⋯Solution (Continue…)
The expansion is valid for
Bloom: Understanding 𝒙 𝟗 < 𝟏 −𝟏 < 𝒙 𝟗 < 𝟏 ∴ −𝟗 < 𝒙 < 𝟗
Solution (Continue…)
4. Bloom: Understanding 𝟏 − 𝟐𝒙 𝟏𝟐 = 𝟏 + 𝟏 𝟐 −𝟐𝒙 + 𝟏 𝟐 𝟏 𝟐 − 𝟏 𝟐! −𝟐𝒙 𝟐 + ⋯ = 𝟏 − 𝒙 − 𝟏 𝟐 𝒙 𝟐 + … Put into 𝒙 = 𝟏 𝟏𝟎𝟎 𝟏 − 𝟐𝒙 𝟏 𝟐 𝟏 − 𝟏 𝟏𝟎𝟎 𝟏 𝟐 = 𝟏 − 𝟏 𝟏𝟎𝟎 − 𝟏 𝟐 𝟏 𝟏𝟎𝟎 𝟐Solution (Continue…)
Bloom: Understanding 𝟗𝟖 𝟏𝟎𝟎 = 𝟐𝟎𝟎𝟎𝟎 − 𝟐𝟎𝟎 − 𝟏 𝟐𝟎𝟎𝟎𝟎 𝟗𝟖 𝟏𝟎𝟎 = 𝟏 − 𝟏 𝟏𝟎𝟎 − 𝟏 𝟐𝟎𝟎𝟎𝟎 𝟕 𝟐 = 𝟏𝟗𝟕𝟗𝟗 𝟐𝟎𝟎𝟎 𝟐 = 𝟏𝟗𝟕𝟗𝟗 𝟏𝟒𝟎𝟎𝟎Self-check
Bloom: Applying
1. Expand the following up to the term in . (a)
(b)
2. For which value of are the expansions in Question 1 valid? 𝒙𝟑 𝟏 + 𝒙 𝟏𝟑 𝟏 − 𝒙 𝟑 𝟏 𝟓 𝒙
Self-check
Bloom: Applying
3. Expand the function in a series of
ascending powers of as far as the term in . State the value of for which each
expansion is valid.
4. Obtain the expansion of up to the term. By putting , find a rational
approximation for . 𝟖 + 𝒙 𝟏𝟑 𝒙 𝒙𝟑 𝟏 + 𝒙 𝒙𝟐 𝒙 = 𝟖 𝟏𝟎𝟎 𝟑
Answer Self-check
(1) (a) (b) (2) (a) (b) Bloom: Applying 𝟏 + 𝟏 𝟑 𝒙 − 𝟏 𝟗𝒙 𝟐 + 𝟓 𝟖𝟏 𝒙 𝟑 𝟏 − 𝟏 𝟏𝟓 𝒙 − 𝟐 𝟐𝟐𝟓𝒙 𝟐 − 𝟐 𝟏𝟏𝟐𝟓 𝒙 𝟑 −𝟏 < 𝒙 < 𝟏 −𝟑 < 𝒙 < 𝟑Answer Self-check
(3) (4) Bloom: Applying 𝟐 + 𝒙 𝟏𝟐 − 𝟏 𝟐𝟖𝟖 𝒙 𝟐 + 𝟓 𝟐𝟎𝟕𝟑𝟔 𝒙 𝟑 + … −𝟖 < 𝒙 < 𝟖 𝟑 = 𝟒𝟑𝟑 𝟐𝟓𝟎BFF’s Technique
Bracket
Fill in the blank
Find value
n positive integer n negative integer or fraction
Formula
BFF’s
technique • 3 terms if power to 2
• 4 terms if power to 3 and so on… before doing expansion.Must in the form of
Step 1: Bracket Step 2: Fill in the blank Step 3: Find value
BFF’s Technique
𝒙 + 𝒂 𝒏 = 𝒌=𝟎 𝒏 𝒏 𝒌 𝒙 𝒌𝒂𝒏−𝒌 𝟏 + 𝒏𝒙 +𝒏(𝒏 − 𝟏) 𝟐! 𝒙 𝟐 +𝒏(𝒏 − 𝟏)(𝒏 − 𝟐) 𝟑! 𝒙 𝟑+ … 𝟐𝒙 + 𝟓 𝟐 𝟏 + 𝒙 −𝟐 + + 𝟏 + 𝟏! 𝟏+ 𝟐! 𝟐+ 𝟑! 𝟑+ ⋯ 𝒂 + 𝒃𝒙 𝒏= 𝒂𝒏 𝟏 +𝒃 𝒂 𝒏 𝟏 + 𝒙 𝒏 𝟐 𝟎 𝟐𝒙 𝟐 𝟓 𝟎+ 𝟐 𝟏 𝟐𝒙 𝟏 𝟓 𝟏+ 𝟐 𝟐 𝟐𝒙 𝟎 𝟓 𝟐 𝟏 + −𝟐 𝟏! 𝒙 𝟏+ −𝟐 −𝟐 − 𝟏 𝟐! 𝒙 𝟐+ −𝟐 −𝟐 − 𝟏 −𝟐 − 𝟐 𝟑! 𝒙 𝟑+ ⋯ 𝟒𝒙𝟐 + 𝟐𝟎𝒙 + 𝟐𝟓 𝟏 − 𝟐𝒙 + 𝟑𝒙𝟐 − 𝟒𝒙𝟑+ ⋯Summary
Bloom: Remembering Binomial expansion Power, n is a positive integer 𝒂 + 𝒃 𝒏 = 𝒏𝟎 𝒂𝒏 + 𝒏𝟏 𝒂𝒏−𝟏𝒃 + 𝒏𝟐 𝒂𝒏−𝟐𝒃𝟐 + 𝒏 𝟑 𝒂 𝒏−𝟑𝒃𝟑 +… 𝑻𝒓+𝟏 = 𝒏𝑪𝒓𝒂𝒏−𝒓𝒃𝒓 Power, n is a negative integer or fraction 𝟏 + 𝒙 𝒏 = 𝟏 + 𝒏𝒙 +𝒏(𝒏 − 𝟏)𝒙 𝟐 𝟐! + 𝒏(𝒏−𝟏)(𝒏−𝟐)𝒙𝟑 𝟑! + … 𝒂 + 𝒃𝒙 𝒏= 𝒂𝒏 𝟏 +𝒃 𝒂 𝒏before applying the expansion of as shown above. 𝟏 + 𝒙 𝒏
