linoluv
Answered

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

Sagot :

semsee45

Answer:

(i)  [tex]64+192 x^2+240 x^4+160x^6[/tex]

(ii) 1072

Step-by-step explanation:

Part (i)

Using Binomial series formula:

[tex](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[/tex]

              [tex]=64+6 \cdot 32\cdot x^2+15 \cdot 16\cdot x^4+20 \cdot 8 \cdot x^6[/tex]

              [tex]=64+192 x^2+240 x^4+160x^6[/tex]

Part (ii)

[tex](1-\frac{3}{x^2})^2=(1-\frac{3}{x^2})(1-\frac{3}{x^2})[/tex]

               [tex]=1-\dfrac{6}{x^2}+\dfrac{9}{x^4}[/tex]

[tex](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)[/tex]

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

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

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

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