Dabour Hub EGEgyptian Curriculum 2026/2027
1-2 The term containing xk and the term free of x

1-2 The term containing xk and the term free of x

Point 5 of 8 — Lesson 1: Counting Principle, Permutations (nPr) and Combinations (nCr)Question bank: 5 questions (5 with pictures)

05 1-2 The term containing xk and the term free of x

Method

  • Writing the whole expansion to find one term is slow, so use the general term.
  • Step 1: write Tr+1 in terms of r.
  • Step 2: add the powers of x in Tr+1 (an expression in r) and put the sum equal to k.
  • Step 3: solve for r. If r is a natural number (and 0≤r≤n), the required term is Tr+1.
  • If r is not a natural number, there is no term containing xk.
  • For the term free of x, put the sum of the powers of x equal to 0.
✍️ Worked examples

Example 11: In the expansion of (x2+2x)9 find the coefficient of x6, the term free of x, and show that there is no term containing x7.

Solution:

  • Tr+1=9Cr (x2)9−r(2x)r=9Cr 2r x18−3r.
  • Coefficient of x6: 18−3r=6, so r=4 and the term is T5 with coefficient 9C4×24=126×16=2016.
  • Free of x: 18−3r=0, so r=6 and the term is T7=9C6×26=84×64=5376.
  • For x7: 18−3r=7 gives r=113, which is not a natural number.
  • So the expansion has no term containing x7.

Equal middle terms

  • To find the value of a quantity that makes two middle terms equal, write both middle terms with the general term and equate them.
✍️ Worked examples

Example 13: In the expansion of (2+x3)9 find x if the two middle terms are equal.

Solution:

  • n=9 is odd, so the middle terms are T5 and T6.
  • T5=9C4 (2)5(x3)4 and T6=9C5 (2)4(x3)5.
  • 9C4=9C5, so equate: 25(x3)4=24(x3)5.
  • Divide by 24(x3)4: 2=x3, so x=6.

📝 Question bank for this point

5 questions linked to this point only, 5 of them with pictures.