Question

i) Find the first 4 terms in the expansion of
$(2 + x ^{2})^{6}$
in ascending powers of x.
ii) Find the term independent of x in the expansion of
$(2 + x ^{2})^{6}(1 - \frac{3}{x {}^{2} } ) {}^{2}$
​

Answer 1

Answer:

(i)  $64+192 x^2+240 x^4+160x^6$

(ii) 1072

Step-by-step explanation:

Part (i)

Using Binomial series formula:

$(2+x^2)^6=2^6+6C1 \cdot 2^{6-1}\cdot x^2+6C2 \cdot 2^{6-2}\cdot(x^2)^2+6C3 \cdot 2^{6-3}\cdot(x^2)^3$

              $=64+6 \cdot 32\cdot x^2+15 \cdot 16\cdot x^4+20 \cdot 8 \cdot x^6$

              $=64+192 x^2+240 x^4+160x^6$

Part (ii)

$(1-\frac{3}{x^2})^2=(1-\frac{3}{x^2})(1-\frac{3}{x^2})$

               $=1-\dfrac{6}{x^2}+\dfrac{9}{x^4}$

$(2+x^2)^6(1-\frac{3}{x^2})^2=(64+192 x^2+240 x^4+160x^6)\left(1-\dfrac{6}{x^2}+\dfrac{9}{x^4} \right)$

$=64-\dfrac{384}{x^2}+\dfrac{576}{x^4}+192x^2-1152+\dfrac{1728}{x^2}+240x^4-1440x^2+2160+...$

There is no need to keep expanding since the remaining will include the variable.

Therefore, the term independent of x = 64 - 1152 + 2160 = 1072