ii)

The expansion of

{(a + x)}^{2}

is

a^{2} + 2 a x + x^{2}

. Multiply the first three terms of

{(1 + 3 x)}^{6}

found in (i) by

a^{2} + 2 a x + x^{2}

, up to terms in

x^{2}

(you don't need to multiply terms that will result with

x^{3}

or higher)

(1 + 18 x + 135 x^{2} + . . .) (a^{2} + 2 a x + x^{2}) = (a^{2}) (1) + (a^{2}) (18 x) + (a^{2}) (135 x^{2}) + . . . + (2 a x) (1) + (2 a x) (18 x) + . . . + (x^{2}) (1) + . . .

(The line above is really just shown for demonstration purposes. In reality we'd probably go straight to the line below)

(1 + 18 x + 135 x^{2} + . . .) (a^{2} + 2 a x + x^{2}) = a^{2} + 18 a^{2} x + 135 a^{2} x^{2} + . . . + 2 a x + 36 a x^{2} + . . . + x^{2} + . . .

We can equate this to

4 + 68 x + b x^{2}

a^{2} + 18 a^{2} x + 135 a^{2} x^{2} + . . . + 2 a x + 36 a x^{2} + . . . + x^{2} + . . . = 4 + 68 x + b x^{2}

The only constant term on the left hand side is "

a^{2}

" therefore equating the constants gives

a^{2} = 4 a = \pm 2

[1]

Only one of these can be correct. Examining the terms in

x

:

\begin{array}{rcl} a & = & - 2, 18 a^{2} x + 2 a x = 18 {(- 2)}^{2} x + 2 (- 2) x = 68 x \\ a & = & 2, 18 a^{2} x + 2 a x = 18 {(2)}^{2} x + 2 (2) x = 76 x \end{array}

But we know that the term in

x

is

68 x

therefore

a \neq 2

and

a = - 2

[1]

Now substitute

a = - 2

into the terms in

x^{2}

\begin{array}{rcl} 135 {(- 2)}^{2} x^{2} + 36 (- 2) x^{2} + x^{2} & = & b x^{2} \\ 540 x^{2} - 72 x^{2} + x & = & b x^{2} \\ 469 x^{2} & = & b x^{2} \end{array}

Therefore

b = 469

[1]

Binomial Theorem (Cambridge O Level Additional Maths)

Exam Questions