linoluv
Answered

Westonci.ca is the best place to get answers to your questions, provided by a community of experienced and knowledgeable experts. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.

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

Thanks for stopping by. We are committed to providing the best answers for all your questions. See you again soon. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.