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WAEC 2012 · Paper 2 · Q9

  1. (a)

    The first three terms of the expansion of (1+mx)n(1 + mx)^n in ascending powers of xx are 1+14x+84x21 + 14x + 84x^2. Find the values of mm and nn.

    Separate values with commas, e.g. 3, −2

  2. (b)

    Using the values of mm and nn obtained in (a), calculate, correct to three significant figures, the value of (1.06)n(1.06)^n.

Worked solution (try it first)

(a)

  1. (1+mx)n=1+n(mx)+n(n−1)2(mx)2+…(1 + mx)^n = 1 + n(mx) + \dfrac{n(n - 1)}{2}(mx)^2 + \ldots
  2. Match the xx terms: nm=14nm = 14, so m=14nm = \dfrac{14}{n}.
  3. Match the x2x^2 terms: n(n−1)2m2=84\dfrac{n(n - 1)}{2}m^2 = 84.
  4. Substitute mm: n(n−1)2×196n2=84\dfrac{n(n - 1)}{2} \times \dfrac{196}{n^2} = 84.
  5. Simplify: 98(n−1)n=84\dfrac{98(n - 1)}{n} = 84, so 98n−98=84n98n - 98 = 84n, and 14n=9814n = 98.
  6. So n=7n = 7 and m=2m = 2.

(b)

  1. (1.06)7=(1+2x)7(1.06)^7 = (1 + 2x)^7 with 2x=0.062x = 0.06, so x=0.03x = 0.03.
  2. (1+2x)7=1+14x+84x2+280x3+…(1 + 2x)^7 = 1 + 14x + 84x^2 + 280x^3 + \ldots
    =1+0.42+0.0756+0.00756+…= 1 + 0.42 + 0.0756 + 0.00756 + \ldots
  3. That is 1.503…1.503\ldots, so (1.06)7=1.50(1.06)^7 = 1.50 to three significant figures.

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