Dabour Hub EGEgyptian Curriculum 2026/2027
1-1 (b) The general term

1-1 (b) The general term

Point 3 of 8 — Lesson 1: Counting Principle, Permutations (nPr) and Combinations (nCr)Question bank: 5 questions (5 with pictures)

03 1-1 (b) The general term

Rule

  • The general term is the term of order r+1 in the expansion of (x+y)n, with 0≤r≤n: Tr+1=nCr  x n−r y r
  • The coefficient of Tr+1 is nCr×(coefficient of the first term)n−r×(coefficient of the second term)r, including signs.
✍️ Worked examples

Example 5: In the expansion of (x3+2x)8 in descending powers of x, find the sixth term and its coefficient.

Solution:

  • Sixth term means r+1=6, so r=5.
  • T6=8C5 (x3)3(2x)5.
  • =56×x9×32 x−5=1792 x4.
  • The coefficient is 1792.

Term counted from the end

  • The term of order m from the end in an expansion with n+1 terms is the term of order n+1−m+1=n−m+2 from the beginning.
✍️ Worked examples

Example 6: Find the fourth term from the end in the expansion of (2x−1x)7.

Solution:

  • The expansion has 8 terms, so the order from the beginning is 8−4+1=5.
  • T5=7C4 (2x)3(−1x)4.
  • =35×8x3×x−4=280 x−1.

Sum and difference of two expansions

  • The expansions of (x+a)n and (x−a)n differ only in the signs of the terms of even order, so: (x+a)n+(x−a)n=2(T1+T3+T5+⋯ ) (x+a)n−(x−a)n=2(T2+T4+T6+⋯ )
  • where T1,T2,… are the terms of (x+a)n.
✍️ Worked examples

Example 7: Simplify (x+2)6+(x−2)6.

Solution:

  • The sum equals 2(T1+T3+T5+T7).
  • =2(x6+6C2x4(2)2+6C4x2(2)4+26).
  • =2(x6+60x4+240x2+64).
  • =2x6+120x4+480x2+128.

Recognising an expansion in disguise

  • An expression such as (3+x)n−nC1(3+x)n−1(1−2x)+⋯ is the expansion of [(3+x)−(1−2x)]n, which simplifies before you pick a term.

Finding unknowns from coefficients

  • Write the coefficients as combinations, then divide one equation by the other to cancel the unknown.
✍️ Worked examples

Example 8: If (1+cx)n=1+20x+a2x2+a3x3+⋯ and 16a2=3a3, find n and c (where c≠0).

Solution:

  • Coefficient of x: nC1 c=20, so nc=20 (1).
  • a2=nC2 c2 and a3=nC3 c3.
  • 16 nC2 c2=3 nC3 c3, so 16 nC2=3 nC3 c.
  • Use nC3nC2=n−23: 16=(n−2) c (2).
  • From (2): nc−2c=16; with (1): 20−2c=16, so c=2.
  • Then n=202=10.

📝 Question bank for this point

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