Find the ratio of the coefficients of the 4th and 6th terms in the expansion of (1 + 2x)^16
You had a similar question like this before
http://www.jiskha.com/display.cgi?id=1375847416
anyway...
(1+2x)^16
= 1 + C(16,1) (2x)^1 + C(16,2) (2x)^2 + ...
term 4 = C(16,3) (2x)^3 = 4480x^3
term 6 = C(16,5) (2x)^5 = 139776x^5
4480 : 139776 = 5:156
To find the ratio of the coefficients of the 4th and 6th terms in the expansion of (1 + 2x)^16, we can use the binomial expansion formula.
The binomial expansion formula is given by:
(x + y)^n = C(n, 0) * x^n * y^0 + C(n, 1) * x^(n-1) * y^1 + C(n, 2) * x^(n-2) * y^2 + ...
where C(n, r) is the binomial coefficient given by:
C(n, r) = n! / (r! * (n-r)!)
In this case, x = 1 and y = 2x, so we can rewrite the formula as:
(1 + 2x)^16 = C(16, 0) * 1^16 * (2x)^0 + C(16, 1) * 1^15 * (2x)^1 + C(16, 2) * 1^14 * (2x)^2 + ...
To find the ratio of the coefficients of the 4th and 6th terms, we need to calculate C(16, 3) and C(16, 5).
C(16, 3) = 16! / (3! * (16-3)!) = 16! / (3! * 13!) = (16 * 15 * 14) / (3 * 2) = 560
C(16, 5) = 16! / (5! * (16-5)!) = 16! / (5! * 11!) = (16 * 15 * 14 * 13 * 12) / (5 * 4 * 3 * 2) = 4368
Therefore, the ratio of the coefficients of the 4th and 6th terms is 560:4368, which can be simplified to 1:7.8.
To find the ratio of the coefficients of the 4th and 6th terms in the expansion of (1 + 2x)^16, we use the formula for the binomial theorem.
The binomial theorem states that for any positive integer n and real numbers a and b,
(a + b)^n = C(n,0)*a^n*b^0 + C(n,1)*a^(n-1)*b^1 + C(n,2)*a^(n-2)*b^2 + ... + C(n,r)*a^(n-r)*b^r + ... + C(n,n)*a^0*b^n,
where C(n,r) is the binomial coefficient, also known as "n choose r", given by the formula:
C(n,r) = n! / (r! * (n-r)!),
and the factorial is denoted by the exclamation mark (!).
Let's apply the binomial theorem to expand (1 + 2x)^16:
(1 + 2x)^16 = C(16,0)*1^16*(2x)^0 + C(16,1)*1^15*(2x)^1 + C(16,2)*1^14*(2x)^2 + ... + C(16,4)*1^12*(2x)^4 + ... + C(16,16)*1^0*(2x)^16.
We can see that the 4th term in the expansion will be C(16,4)*1^12*(2x)^4, and the 6th term will be C(16,6)*1^10*(2x)^6.
So, the ratio of the coefficients of the 4th and 6th terms is:
C(16,4)*1^12*(2x)^4 / C(16,6)*1^10*(2x)^6.
Now, let's calculate these binomial coefficients:
C(16,4) = 16! / (4! * (16-4)!) = 16! / (4! * 12!) = (16 * 15 * 14 * 13) / (4 * 3 * 2 * 1) = 1820,
C(16,6) = 16! / (6! * (16-6)!) = 16! / (6! * 10!) = (16 * 15 * 14 * 13 * 12 * 11) / (6 * 5 * 4 * 3 * 2 * 1) = 8008.
Plugging these values back into the ratio, we get:
1820 * 1^12 * (2x)^4 / 8008 * 1^10 * (2x)^6 = 1820 * (2x)^4 / 8008 * (2x)^6,
Simplifying this expression, we get:
3640x^4 / 16016x^6,
Finally, canceling out the common factor x^4, we have:
3640 / 16016x^2.
Therefore, the ratio of the coefficients of the 4th and 6th terms in the expansion of (1 + 2x)^16 is 3640 / 16016.