Coefficients

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To find the coefficient of u^2 v^9 in the expansion of (2u - 3v + u^2 - v^2)^9, we need to use the binomial theorem to expand the expression and then identify the term that contains u^2 v^9.

The binomial theorem states that the expansion of (a + b)^n can be written as:

(a + b)^n = ∑(k=0 to n) [n choose k] a^(n-k) b^k

where [n choose k] is the binomial coefficient, given by:

[n choose k] = n! / (k! (n-k)!)

Using this formula, we can expand (2u - 3v + u^2 - v^2)^9 as:

(2u - 3v + u^2 - v^2)^9 = ∑(k=0 to 9) [9 choose k] (2u)^(9-k) (-3v)^k (u^2)^{9-k} (-v^2)^k

= ∑(k=0 to 9) [9 choose k] 2^(9-k) (-3)^k u^(18-2k) v^k u^(2k) (-1)^k v^(2k)

= ∑(k=0 to 9) [9 choose k] 2^(9-k) (-3)^k (-1)^k u^18 v^k

= [9 choose 9] 2^0 (-3)^9 (-1)^9 u^18 v^9 + [9 choose 7] 2^2 (-3)^7 (-1)^7 u^14 v^7 + ...

We are interested in the coefficient of u^2 v^9, which appears only in the first term of the expansion. Therefore, the coefficient is given by:

[9 choose 9] 2^0 (-3)^9 (-1)^9 = 1 * (-19683) * (-1) = 19683

Therefore, the coefficient of u^2 v^9 in the expansion of (2u - 3v + u^2 - v^2)^9 is 19683. You can also visit  Check My Rota Login

114th term   (2u - 3v + u^2 - v^2)^9== - 577368 u ^2 v^9

Find the coefficient of u^2 v^9 in the expansion of (2u - 3v + u^2 - v^2)^9

[u(2+u)+v(-3-v) ]^9

$\binom{9}{n}\;[u(u+2)]^n\;[-v(v+3)]^{9-n}\\~\\ \binom{9}{n}\;[u^n(u+2)^n\;][-v^{9-n}(v+3)^{9-n}]\\~\\ (-1)^{n-1}\;\binom{9}{n}\;[u^n(u+2)^n\;][v^{9-n}(v+3)^{9-n}]\\~\\ (-1)^{n-1}\;\binom{9}{n}\;[u^n(u+2)^n\;][v^{9-n}(v+3)^{9-n}]\\~\\$

To get u^2   n must be 1

$(-1)^{9-1}\;\binom{9}{1}\;[u^1(u+2)^1\;][v^{9-1}(v+3)^{9-1}]\\~\\ =(-1)^{8}\;*9*\;[u(u+2)\;][v^{8}(v+3)^{8}]\\~\\ =9v^8\;[u(u+2)\;][(v+3)^{8}]\\~\\ $

The constant term for  (v+3)^8 = 3^8

So the u^2v^8 term will be in here

$9v^8\;[u(u+2)\;]3^8\\~\\ 9v^8\;[u^2+2u\;]3^8\\~\\ The \;\;u^2v^8 \text{ term will be}\\~\\ 9*3^8*u^2v^8 = 59049\;u^2v^8$

LaTex

\binom{9}{n}\;[u(u+2)]^n\;[-v(v+3)]^{9-n}\\~\\
\binom{9}{n}\;[u^n(u+2)^n\;][-v^{9-n}(v+3)^{9-n}]\\~\\
(-1)^{n-1}\;\binom{9}{n}\;[u^n(u+2)^n\;][v^{9-n}(v+3)^{9-n}]\\~\\
(-1)^{n-1}\;\binom{9}{n}\;[u^n(u+2)^n\;][v^{9-n}(v+3)^{9-n}]\\~\\

(-1)^{9-1}\;\binom{9}{1}\;[u^1(u+2)^1\;][v^{9-1}(v+3)^{9-1}]\\~\\
=(-1)^{8}\;*9*\;[u(u+2)\;][v^{8}(v+3)^{8}]\\~\\

9v^8\;[u(u+2)\;]3^8\\~\\
9v^8\;[u^2+2u\;]3^8\\~\\
The \;\;u^2v^8 \text{ term will be}\\~\\
9*3^8*u^2v^8 = 59049\;u^2v^8