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You have seen that the total number of lights in a triangular lighting rig is given by T(n) = n2/2 + n/2, while T(n+1) = n2/2 + 3n/2 + 1. If T(n) times T(n+1) = f(n), what is the reduced form of f(n)?
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u shud be able to do this. factorize T(n) and T(n+1) and then multiply.
g(n) = g(n-1) + n?
T(n) = n(n+1)/2 T(n+1) = (n+1)(n+2)/2 T(n)*T(n+1) = f(n) so multiply T(n) and T(n+1) and calculate f(n)...
g(n) = g(n -1) + 1 ?
where did you get g(n) from?
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when i solved for it
oh i was on wrong problem
i got f(n)=n^4=n^2/2+n^2/2+n/2
its not correct. when u multiply T(n) and T(n+1) u get f(n) = n(n+1)^2(n+2)/4 U can expand it and arrive at f(n) = (n^4+4n^3+5n^2+2n)/4. Its just simple multiplication.
didi u ever figure out the awnser
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