Find the indicated term in the expansion of the given binomial.
The term that does not contain x in the expansion of
(6x+1/2x)^12
Is there an easier way to find the answer instead of going through finding 13 terms
Thanks.
When 'x' is not in the term, than x will have the same power in the numerator and denominator, and will hence cancel out.
So, (6x) and (1/2x) will be raised to same power.
Which term will this happen in?
(a+b)^12
where is power of a same as of b
C(12,6) a^6 b^6
where b is binomial coef
C(12,6) = 12!/[6! 6!] = 12*11*10*9*8*7/6!
=2*11*3*2*7
=924
so
924[(6x)^6 * 1/(2x)^6 ]
924 * 6^6/2^6
the 11th term? x can't be cancelled
Thanks Damon
the "center or middle" term ... 7th
Yes, there is a faster way to find the term that does not contain x in the expansion of (6x + 1/2x)^12 without going through finding all 13 terms.
In this expression, the term that does not contain x will only arise from the constant term when multiplying the binomial multiple times.
The constant term arises when we choose the constant term, 1/2x, from one factor and the constant term, 1/2x, from the other factor in each term of the expansion.
To find the term that does not contain x, we need to find the coefficient of the constant term. The constant term will be x raised to the power of 0, so we need to find the coefficient of x^0.
To find the coefficient of x^0 in the expansion, we can use the binomial theorem. According to the binomial theorem, the coefficient of x^r in the expansion of (a + b)^n is given by the binomial coefficient formula:
C(n, r) = n! / (r! * (n-r)!)
In this case, n = 12 and r = 0.
C(12, 0) = 12! / (0! * (12-0)!)
Since 0! = 1, we can simplify the expression further:
C(12, 0) = 12! / (1 * 12!)
The factorials in the numerator and denominator cancel out, leaving us with:
C(12, 0) = 1
Therefore, the term that does not contain x in the expansion of (6x + 1/2x)^12 is simply 1.