Binomial expansion for positive integer powers, binomial coefficients, approximations.
Answer at least 3 of 3 correctly to complete this section.
You already know how to expand (a+b)2=a2+2ab+b2. You might even know (a+b)3. But what about (1+2x)8? Multiplying out eight brackets by hand would take forever.
The binomial expansion gives you a formula that works for any positive integer power. It is one of the most powerful algebraic tools in A-Level Mathematics, appearing in approximation problems, probability (the binomial distribution gets its name from this), and series work.
Each row of Pascal’s triangle gives the coefficients of a binomial expansion:
| Power | Coefficients |
|---|---|
| (a+b)0 | 1 |
| (a+b)1 | 11 |
| (a+b)2 | 121 |
| (a+b)3 | 1331 |
| (a+b)4 | 14641 |
Each entry is the sum of the two entries above it. For larger powers, we use the formula:
(rn)=r!(n−r)!n!
This is also written nCr or C(n,r).
(a+b)n=∑r=0n(rn)an−rbr
=(0n)an+(1n)an−1b+(2n)an−2b2+⋯+(nn)bn
The power of a decreases from n to 0; the power of b increases from 0 to n. The powers always add up to n.
Find the first 4 terms in ascending powers of x of (1+2x)8.
Here a=1, b=2x, n=8. The general term is (r8)(1)8−r(2x)r=(r8)(2x)r.
r=0:(08)(2x)0=1
r=1:(18)(2x)1=8×2x=16x
r=2:(28)(2x)2=28×4x2=112x2
r=3:(38)(2x)3=56×8x3=448x3
(1+2x)8=1+16x+112x2+448x3+⋯
The most common error: Using (rn)×xr instead of (rn)×(2x)r. You must raise the entire second term to the power r. So (2x)2=4x2, not 2x2.
Answer at least 3 of 3 correctly to complete this section.
To find the coefficient of xk, identify which value of r gives xk in the general term, then compute that term.
Find the coefficient of x3 in the expansion of (3−x)7.
The general term is:
(r7)(3)7−r(−x)r=(r7)⋅37−r⋅(−1)r⋅xr
For x3: set r=3.
(37)×34×(−1)3=35×81×(−1)=−2835
Watch the sign. When the second term is negative (like −x), the sign alternates: (−x)1=−x, (−x)2=+x2, (−x)3=−x3, etc.
In the expansion of (1+kx)6, the coefficient of x2 is 60. Given k>0, find k.
Coefficient of x2:
(26)k2=15k2
Setting equal to 60:
15k2=60⟹k2=4⟹k=2
(since k>0).
Answer at least 3 of 3 correctly to complete this section.
The term independent of x is the constant term — the one where the power of x is zero.
Find the term independent of x in the expansion of (2x+x23)6.
Write the general term:
(r6)(2x)6−r(x23)r=(r6)⋅26−r⋅3r⋅x6−r⋅x−2r
=(r6)⋅26−r⋅3r⋅x6−3r
For the term independent of x: set the power of x to zero.
6−3r=0⟹r=2
Substitute r=2:
(26)×24×32=15×16×9=2160
Check that r is a whole number between 0 and n. If it is not, there is no term independent of x.
Answer at least 3 of 3 correctly to complete this section.
In the expansion of (2+x)5, find the coefficient of x3.
General term: (r5)⋅25−r⋅xr.
For x3: r=3.
(35)×22×x3=10×4×x3=40x3
The coefficient is 40.
Use the first three terms of the expansion of (1+x)10 to approximate 1.0210.
(1+x)10≈1+10x+45x2+⋯
Set x=0.02:
1.0210≈1+10(0.02)+45(0.0004)
=1+0.2+0.018=1.218
(The exact value is 1.21899…, so the approximation is excellent.)
| Question type | What to do |
|---|---|
| ”Find the first 4 terms” | Expand with r=0,1,2,3 |
| “Find the coefficient of xk“ | Set r=k (or solve for r if terms have mixed powers) |
| “Find the term independent of x“ | Set total power of x=0, solve for r |
| “Given the coefficient is…, find k“ | Write coefficient in terms of k, set equal, solve |
Answer at least 3 of 4 correctly to complete this section.
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