Solution of middle terms expansion of (2x^2+1/x)^9
the middle term would be term(5)
term(5) = C(9,4) (2x^2)^5 (1/x)^4
= 126 (32x^10) (1/x^4)
= 4032 x^6
actually, since there are 10 terms, and the question asked for the middle terms, we have to include the next term:
C(9,5) (2x^2)^4 (1/x)^5
= 126 (16x^8) (1/x^5)
= 2016 x^3
yup, Steve is correct
To find the solution of the middle terms expansion of (2x^2 + 1/x)^9, you can use the Binomial Theorem.
The Binomial Theorem is a formula that allows us to expand expressions of the form (a + b)^n, where a and b are any numbers, and n is a positive integer.
The formula for the Binomial Theorem is:
(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, n-1) * a^1 * b^(n-1) + C(n, n) * a^0 * b^n
In this case, we have (2x^2 + 1/x)^9.
To find the middle terms, we need to expand the powers of x^2 and 1/x in the expression.
First, let's rewrite (2x^2 + 1/x)^9 as (2x^2)^9 * (1/x)^9.
Expanding (2x^2)^9 using the Binomial Theorem, we get:
C(9, 0) * (2x^2)^9 * (1/x)^0 + C(9, 1) * (2x^2)^8 * (1/x)^1 + C(9, 2) * (2x^2)^7 * (1/x)^2 + ... + C(9, 8) * (2x^2)^1 * (1/x)^8 + C(9, 9) * (2x^2)^0 * (1/x)^9
Simplifying each term, we get:
1 * 2^9 * x^18 * 1 + 9 * 2^8 * x^16 * 1/x + 36 * 2^7 * x^14 * 1/x^2 + ... + 9 * 2^1 * x^2 * 1/x^8 + 1 * 2^0 * 1 * 1/x^9
This can be further simplified to:
2^9 * x^18 + 9 * 2^8 * x^15 + 36 * 2^7 * x^12 + ... + 9 * 2^1 * x^-6 + 2^0 * x^-9
So, the solution to the middle terms expansion of (2x^2 + 1/x)^9 is:
2^9 * x^18 + 9 * 2^8 * x^15 + 36 * 2^7 * x^12 + ... + 9 * 2^1 * x^-6 + 2^0 * x^-9