Determine the constant term in.https://webwork.math.su.se/webwork2_...1d3f2168b1.png
idk how to solve it :(
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Determine the constant term in.https://webwork.math.su.se/webwork2_...1d3f2168b1.png
idk how to solve it :(
each term in a binomial expansion is of the form ...
In this case,... the value of
will be such that the variable factors cancel.
i did not get it, i dont think i really understand what they want me to answer, with other words i maybe dont understand problem... im confused
1. have you learned the binomial theorem?Quote:
Originally Posted by Petrus
2. do you understand that a constant term has no variable factors?
hi!:)Quote:
Originally Posted by skeeter
We have not come here yet, but I want to study in advance. I did read but did not understand clearly i dont like my book :(
your book is not the sole source of knowledge. seek and ye shall find ...Quote:
Originally Posted by Petrus
Binomial Theorem (part 1) | Precalculus | Khan Academy
ok i did watch and i kinda understand but idk how in my problem to get out the constant term :S
I showed you how in post #2.
A constant terms comes out when a^n-k * b^k is a constant, and does not involve any variables. In your case, the binomial expansion has only one variable, x. That means, in (3x^6)^n - k * (5/x)^k does not contain any x. For it to happen, when n = 21, you need to equate the power of x to zero, since x^0 = 1, is a constant, and other powers yield terms with x, and so, not a constant term.
(x^6)^21 - k * (1/x)^k = x^0
(x^6)^21 - k * (x^-1)^k = x^0
(x^6)^21 - k * x^-k = x^0
x^(6(21-k) - k) = x^0
Equating coefficients, you get k = 18, which is what he used to get the constant term above.
Salahuddin
Maths online
I solved this problem :) forgot to say it ty anyway :)