IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: halosh92 on February 27, 2010, 07:27:17 am
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(a) Expand (2 + y)^6 in ascending powers of y as far as the term in y^3, simplifying
each coefficient.
(b) Hence expand (2 + x ? x^2)^6 in ascending powers of x as far as the term in x^3,
simplifying each coefficient.
i know how to do the (a) part but am stuck at part (b)
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Can you restate the question?
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Can you restate the question?
how do u do the second part?
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I meant this (2 + x ? x^2)^6
with the question mark in the middle
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I meant this (2 + x ? x^2)^6
with the question mark in the middle
sorry typing mistake its:
(2+ x- x^2)^6
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Ok will do it when I get home
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(2+x)^6=1*2^6*x^0+6*2^5*x^1+15*2^4*x^2+20*2^3*x^3+....
=64+192x+240x^2+160x^3....
Now replace all the x's by x-x^2
64+192(x-x^2)+240(x-x^2)^2 +160(x-x^2)^3
64-192x-192x^2+240(x^2-2x^3)+160x^3 ignoring all powers of x higher than x^3
=64+192x+48x^2-320x^3
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(2+x)^6=1*2^6*x^0+6*2^5*x^1+15*2^4*x^2+20*2^3*x^3+....
=64+192x+240x^2+160x^3....
Now replace all the x's by x-x^2
64+192(x-x^2)+240(x-x^2)^2 +160(x-x^2)^3
64-192x-192x^2+240(x^2-2x^3)+160x^3 ignoring all powers of x higher than x^3
=64+192x+48x^2-320x^3
ok thankyou!!!! :D ;)