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    Okay, we had this question in a test and I was just wondering how would you do this;

    1. Expand (2+y)^6 upto and including y^3 co-efficient.
    2. Hence or otherwise expand (2-x-x^2)^6 upto and including the x^3 co-efficient.
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    There's a few ways you could do this, either brute force expanding term by term and picking out everything up to y^3, or you can use Pascal's triangle, or the binomial formula from (amongst other places) wikipedia .

    Probably best to do the brute force method for lower powers yourself to be confident that the formula works, but for powering up 6 times, you're best off using the formula and your calculator

    For the second part, can you see how you could use the first part to get the answer plus some extra terms? You can easily chop off the extra terms after expanding.
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    (Original post by mintmocha)
    Okay, we had this question in a test and I was just wondering how would you do this;

    1. Expand (2+y)^6 upto and including y^3 co-efficient.
    2. Hence or otherwise expand (2-x-x^2)^6 upto and including the x^3 co-efficient.
    \displaystyle (a+b)^n=\sum^n_{i=0} \binom{n}{i}a^{n-i}b^{i}
    for 1. n=6 and upto i=3
    for example coefficient of y^3 \dysplaystyle \binom{6}{3}2^3=20\cdot 8=160

    for 2.

    Consider 2-x as 'a' and -x^2 as 'b' and use the identity.
 
 
 
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