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Using the binomial theorem we can expand (x^2+1/x)^9
=>x^18 + 9x^16*(1/x) + 36x^14*(1/x^2) + 84x^12*(1/x^3) + 126x^10*(1/x^4) + 126x^8*(1/x^5) + 84x^6*(1/x^6)+ 36x^4*(1/x^7) + 9*x^2*(1/x^8) + 1/x^9.
Therefore the term that does not have x is the one highlighted in the series above. Here only 84 is left as x in the first term which appears in the numerator is cancelled by the x in the second term which appears in the denominator.
The term with an x is the 7th term of the binomial expansion.
The binomial expansion of (A+B)^n = a^n+nC1*A^(n-1)*B+nC2*A^(n-2)*B2+....nCr-1*A^(n-r+1)*B^(r-1)+...+B^n.
So the rth term os Tr = nCr-1*A^(n-r+1)*B^(r-1).
In this case A = x^2, B = 1/x. n = 9.
If r is the term which does not have xt, then the exponent of x must be zero.
So Tr = 9Cr -1*(x^2)^(9-r+1)*x^(-1)(r-1)....(1)
So the exponent is 2(9-r+1)-r+1 = 0
18-2r+2-r+1 = 0
21- 3r = 0.
Therefore 21 = 3r, So r = 21/3 = 7.
Therefore T6 = 9C7 = (C2 = 9*8/2 = 36.
Therefore the 7th term in(x^2+x)^9 is 36.
To determine this term, we'll have to write the formula of the term that occupies the place k+1:
T k+1 = C(n,k)*a^(n-k)*b^(k) (*)
We'll identify the terms a and b from the development:
a = x^2 and b = 1/x
We don't know the place that the term occupies, so, we'll have to determine k.
n = 9
We'll substitute all we know in the formula (*);
T k+1 = C(9,k)*x^[2(9-k)]*x^(-k)
We'll add the superscripts of x:
T k+1 = C(9,k)*x^[2(9-k)-k]
Since this term doesn't contain x, we'll impose the constraint that the superscript to be zero, because x^0 = 1.
[2(9-k)-k] = 0
We'll remove the brackets:
18 - 2k - k = 0
We'll combine like terms:
-3k = -18
We'll divide by -3:
k = 6
The term that doesn't contain x occupies the 7th place, so it is the 7th term of development.
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