Middle terms, sums of coefficients and the term containing a given power of x
الأسبوع 2 · الحد الأوسط أو الحدان الأوسطان في المفكوك ومسائل متنوعة على الدرس الأول وإيجاد الحد المشتمل على قوة معينة من س · 3 دروس
3
Finding the term that contains a given power of x
Pure Mathematics Sec 3 Term 1, Unit 1, Lessons 1–2
GoalFind the term that contains a given power of x, or show that there is none.
New words
term containing xᵏ · the term whose power of x equals kwhole number r · r must be a whole number from 0 to n, or the term does not exist
Explanation
To find the term that contains xᵏ, write the general term T(r+1), collect the powers of x into one expression in r, set it equal to k and solve for r. The table shows the five steps. The answer is only good if r is a whole number with 0 ≤ r ≤ n. If r comes out as a fraction or is outside that range, the expansion has no such term, and we say so. When the question asks for a coefficient, read only the number in front of xᵏ after you put r back. A factor outside the bracket, as in x²(1 + 2x)⁶, just shifts the power: the coefficient of x⁵ in the product is the coefficient of x³ inside the bracket. The term that contains x⁰, in other words the term with no x, is the case k = 0; we study it fully next week.
Step
What to do
For (x² + 3/x)⁹ and x⁹
1
Write T(r+1)
9Cr (x²)⁹⁻ʳ (3/x)ʳ
2
Simplify the power of x
9Cr 3ʳ x¹⁸⁻³ʳ
3
Set the power equal to k
18 − 3r = 9
4
Solve for r
r = 3
5
Check r is whole and 0 ≤ r ≤ n, then read the coefficient
9C3 × 3³ = 2268
Worked examples
Find the coefficient of x⁹ in (x² + 3/x)⁹. T(r+1) = 9Cr (x²)⁹⁻ʳ (3/x)ʳ = 9Cr 3ʳ x¹⁸⁻³ʳ. Set 18 − 3r = 9, so r = 3. The coefficient is 9C3 × 3³ = 84 × 27 = 2268.
Is there a term in x⁴ in the same expansion? Set 18 − 3r = 4. Then r = 14/3, which is not a whole number, so there is no term in x⁴. (Only powers 18, 15, 12, …, 0, −3, …, −9 occur: they drop by 3 each time.)
Find the coefficient of x⁵ in x²(1 + 2x)⁶. It equals the coefficient of x³ in (1 + 2x)⁶, that is 6C3 × 2³ = 20 × 8 = 160.
Class activity
Powers on a staircase. For (x² + 3/x)⁹ write the power 18 − 3r for r = 0, 1, 2, … and see which numbers can occur. Then say, without solving, whether x⁶ or x⁷ can appear.
