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secondary 4 | A Maths
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for the top part, I’ve solved till the power of 3. How do we incorporate it into the second part?
Date Posted: 4 years ago
Views: 263
for 1st expansion you should have gotten …
1-6x+15x^2-20x^3+...
for 2nd expansion you replace x with (x-x^2) in the result of 1st expansion …
(1-x+x^2)^6
=[1-(x-x^2)]^6
=1-6(x-x^2)+15(x-x^2)^2-20(x-x^2)^3+...
=1
-6(x-x^2)
+15(x^2-2x^3+...)
-20(x^3-....)
+ ….
… (am assuming that you only need up to x^3 term).
… normally for the 2nd type of expansion, the question will ask for only the 1st few terms so we can disregard higher powers.
1-6x+15x^2-20x^3+...
for 2nd expansion you replace x with (x-x^2) in the result of 1st expansion …
(1-x+x^2)^6
=[1-(x-x^2)]^6
=1-6(x-x^2)+15(x-x^2)^2-20(x-x^2)^3+...
=1
-6(x-x^2)
+15(x^2-2x^3+...)
-20(x^3-....)
+ ….
… (am assuming that you only need up to x^3 term).
… normally for the 2nd type of expansion, the question will ask for only the 1st few terms so we can disregard higher powers.
See 1 Answer
Replace the x above by (x-x²) and sub into all the binomial expansion that you've done.
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