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Post Info TOPIC: A couple NS questions


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A couple NS questions
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I seem to have problems with these questions and can not find any simple way of doing them.


(1) (4 7/12)  * (2 2/5)


(2) (2 2/5) * (5 1/6)   -- (improper fraction)


*(3) 12.75 x 28300 / 102


(4) the expansion of (2x - y)^5 has __ terms


(5) the simplified coefficiant of the 4th term in the expansion of (2x-y)^5


   *or any other find the coefficiant of the nth term in an expansion


That's it for now.


Thank you,


Joshua Lambert



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fht


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1 and 2. There's no way. Turn them into improper fractions. Well, the process would be a lot faster if you notice the relationship between 2 2/5 and /12. But basically, it's improper-ization, and reduction.


3. I'd do 12.75*280 = 12*280 + .75*280 = 3360+210 = 3570


4. The expansion of a binomial to the nth power has n+1 terms


5. Memorize your pascal triangle. On the "5" roll it's 1,5,10,10,5,1. And the x's power on that term is 2 and y's is 3 so it's 10*(2^2)*(-1^3) = -400. The key is to notice the relationship btw Pascal triangle and the coefficients of expansions of binomial to different powers.


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"*(3) 12.75 x 28300 / 102"


12.75 = 12 3/4 = 51/4, making it (51*28300)/(4*102)


51 and 102 cancel to make it 28300 / 8 which is about 3500



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Thank you for your help! This will help me very much.



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(5) the simplified coefficiant of the 4th term in the expansion of (2x-y)^5:

Another way to do this problem (rather than memorizing Pascal's triangle coefficients) is to use combinations:


5C(4-1)

In this case the exponent is 5 and the nth term is 4, so take the combination of 5 items (4-1) or 3 at a time. This equals 10 which is the same as you get using Pascal's triangle. Then multiply this times the square of the (2) coefficient times the cube of the (-1) coefficient (because the 4th term is x2y3) to get your answer (which should be -40).

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fht


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Ay, si.

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