SOLUTION: What is the value of the constant term of {{{(2x^3-1/x^2)^10}}}

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Question 1117810: What is the value of the constant term of %282x%5E3-1%2Fx%5E2%29%5E10
Found 3 solutions by solver91311, greenestamps, ikleyn:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


The 7th term



You can do your own arithmetic

John

My calculator said it, I believe it, that settles it


Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


To get a constant term, you need to take the term with x^3 4 times and the term with (1/x^2) 6 times:
%28x%5E3%29%5E4%2A%281%2Fx%5E2%29%5E6+=+%28x%5E12%29%2A%281%2Fx%5E12%29 which is a constant.

Taking the (2x^3) term 4 times out of 10 and the (-1/x^2) term 6 times gives you a coefficient (and therefore a constant term) of

Answer by ikleyn(52780) About Me  (Show Source):
You can put this solution on YOUR website!
.

            Writing by other tutors leaves you without a guidance on how to get this magic/miracle value  k = 6  (the "seventh term").

            Therefore,  below I placed the full solution with detailed explanations of all auxiliary moments.


The binomial expansion is this formula


%28a%2Bb%29%5En = a%5En + C%5Bn%5D%5E1%2Aa%5E%28n-1%29%2Ab + C%5Bn%5D%5E2%2Aa%5E%28n-2%29%2Ab%5E2 + C%5Bn%5D%5E3%2Aa%5E%28n-3%29%2Ab%5E3 + . . . + C%5Bn%5D%5E%28n-1%29%2Aa%5E1%2Ab%5E%28n-1%29 + b%5En


In our case,  n = 10,  a = 2x%5E3,  b = -1%2Fx%5E2,  therefore, the binomial expansion in out case is  


%282x%5E3-+1%2Fx%5E2%29%5E10 = %282x3%29%5E10 + C%5B10%5D%5E1%2A%282x%5E3%29%5E9%2A%28-1%2Fx%5E2%29 + C%5B10%5D%5E2%2A%282x%5E3%29%5E8%2A%28-1%2Fx%5E2%29%5E2 + C%5B10%5D%5E3%2A%282x%5E3%29%5E7%2A%28-1%2Fx%5E2%29%5E3 + . . . + C%5B10%5D%5E9%2A%282x%5E3%29%2A%28-1%2Fx%5E2%29%5E9 + %28-1%2Fx%5E2%29%5E10.


The common term is  C%5B10%5D%5Ek%2A%282x%5E3%29%5E%2810-k%29%2A%28-1%2Fx%5E2%29%5Ek.


This common term is the constant term  at  3*(10-k) - 2k = 0,  i.e.  30 - 3k - 2k =0,  or  30 = 5k  ====>  k = 30%2F5 = 6 (the seventh term).


Then the coefficient at this term is  C%5B10%5D%5E6%2A2%5E%2810-6%29%2A%28-1%29%5E6 = C%5B10%5D%5E6%2A2%5E4%2A%28-1%29%5E6 = %28%2810%2A9%2A8%2A7%29%2F%281%2A2%2A3%2A4%2A5%2A6%29%29%2A2%5E4 = 3360.

The problems like this one are very standard,  and you can often meet them on exams or on tests.

Therefore,  it is beneficiary for you to have and to know this standard and full solution.


             * * * Memorize it as a mantra ! * * *


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To see other similar solved problems,  look into the lessons
    - Binomial Theorem, Binomial Formula, Binomial Coefficients and Binomial Expansion,  Problem  2
    - Solved problems on binomial coefficients,  Problems  1  and  2
in this site.