The middle term of the expansion of (ax+ay)^n is 14580x^3y^3 , Find the value of a and the value of n, where a,n are natural numbers.
Please explain how you got to that answer
Please explain how you got to that answer
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If the middle term has x^3y^3, then n is 6, since the exponents add to 6
The coefficients on (x+y)^6 are C(6,0); C(6,1); etc
Then the fourth term has a coefficient of C(6,3)= (6!)/(3!*3!)= 20
(ax+ay)^6= (ax)^6+ 6(ax)^5(ay)^1+ ...
Then the fourth term is 20(ax)^3(ay)^3= 20a^3x^3 * a^3y^3= 20a^6* x^3y^3
Now 20a^6= 14580
a^6= 729
a= 729^(1/6)= 3
then n= 6, and a= 3
Hoping this helps!
The coefficients on (x+y)^6 are C(6,0); C(6,1); etc
Then the fourth term has a coefficient of C(6,3)= (6!)/(3!*3!)= 20
(ax+ay)^6= (ax)^6+ 6(ax)^5(ay)^1+ ...
Then the fourth term is 20(ax)^3(ay)^3= 20a^3x^3 * a^3y^3= 20a^6* x^3y^3
Now 20a^6= 14580
a^6= 729
a= 729^(1/6)= 3
then n= 6, and a= 3
Hoping this helps!