الرياضيات البحتة (بالإنجليزية) الصف الثالث الثانوي، الأسبوع 3: The term free of x, and showing that a power of x does not occur

The term free of x, ratios of terms, and the greatest term

الأسبوع 3 · الحد الخالي من س والنسبة بين حدين متتاليين وبين معاملي حدين متتاليين ومعاملات ثلاثة حدود متتالية والحد الأكبر · 3 دروس

The term free of x, and showing that a power of x does not occur

Pure Mathematics Sec 3 Term 1, Unit 1, Lessons 2–3

GoalFind the term free of x, and show that a given power of x cannot occur.

New words

term free of x · the term whose power of x is 0, a plain numberpower equation · the equation that sets the power of x in T(r+1) equal to k

Explanation

A term is free of x when the exponent of x in it is 0. Find it as in last week with k = 0: simplify the power of x in T(r+1), set it equal to 0 and solve for r. For (x² + 1/x)⁶ the power is 2(6 − r) − r = 12 − 3r, and the table shows that it is 0 at r = 4, so the free term is T5 = 6C4 = 15. The same idea answers questions about n. For (x³ + 1/x)ⁿ the power is 3(n − r) − r = 3n − 4r, so a free term needs r = 3n/4 to be a whole number, which happens exactly when n is a multiple of 4. If it is not, the expansion has no free term at all. To show that a power does not occur, solve the power equation and point out that r is not a whole number (or is negative, or bigger than n). Be careful with brackets: for 1/(2x) the 2 is raised to the power r too.

Powers of x in (x² + 1/x)⁶

r	term	power of x
0	T1	12
1	T2	9
2	T3	6
3	T4	3
4	T5	0
5	T6	−3
6	T7	−6

Worked examples

Find the term free of x in (2x² + 1/x)⁹. The power is 2(9 − r) − r = 18 − 3r. Set 18 − 3r = 0, so r = 6. The term is T7 = 9C6 · 2³ = 84 × 8 = 672.

For which n has (x³ + 1/x)ⁿ a term free of x? The power is 3n − 4r = 0, so r = 3n/4, and n must be a multiple of 4. For n = 8: r = 6 and the free term is T7 = 8C6 = 28. For n = 10: r = 7.5, so there is no free term.

The term free of x in (ax + 1/x)⁸ equals the coefficient of x². Find a. T(r+1) = 8Cr a⁸⁻ʳ x⁸⁻²ʳ. Free: r = 4, giving 70a⁴. For x²: r = 3, giving 56a⁵. So 70a⁴ = 56a⁵, and since a ≠ 0, a = 70/56 = 5/4.

Class activity

Staircase of powers. For (x³ + 1/x)⁶ list the powers 3(6 − r) − r for r = 0 to 6. Is 0 on the list? Now do the same for (x³ + 1/x)⁷ and explain what you see.

Practice

Work out the term of (x² − 1/x)⁶ that has no x in it.

Work out the term of (x³ − 1/(2x))⁸ that has no x in it.

For which n has (x² + 1/x³)ⁿ a term free of x? Give the free term when n = 10.

Show that (x³ − 1/x²)⁸ has no term in x⁷.

Find the coefficient of x⁻³ in (x² + 1/x)⁹.