Q#149 - i understand how to do pascals triangle but the rest i dont? Someone help?
I know klutzy is going to say- write the dam question because i dont carry book around all day- here is. What is product of 2nd and 3rd term of expansion of (3x-2y)^5?
Remember that these problems require combinations to figure out individual terms. If the exponent is 'n', then there are 'n+1' terms in the final expansion - from the constant up to the nth power.
Each individual term is computed using a combination and knowledge of what the variables will be. The first term is (n choose 0) x^n y^0. The second term is (n choose 1) x^(n-1) y^1.
In general, the k-th term is given by (n choose [k-1] ) x^(n-k+1) y^(k-1).
If you want the 2nd term then k-1 = 2-1 = 1. You have n = 5.
(5 choose 1) x^4 y^1 = 5x^4y^1. But remember that x is really '3x' and y is really '-2y'.
5(3x)^4 (-2y)^1 = 5(81x^4)(-2y) = -810(x^4)
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If you want the 3rd term then k-1 = 3-1 = 2. You still have n = 5.
(5 choose 2) x^3 y^2 = 10(x^3)(y^2). Again, x is really '3x' and y is really '-2y'.
10(3x)^3(-2y)^2 = 10(27x^3)(4y^2) = 1080(x^3)(y^2)
Now multiply those 2 out and you'll get -874,800(x^7)(y^3).
