mathematical induction
asked Dec 22, 2016 in Other Math Topics by nuel

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2 Answers

http://www.tiger-algebra.com/drill/1.3²_2.3²_3.3²_..._n.3~n=3/4[(2n-1)3~n_1]/

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answered Dec 22, 2016 by Mathical Level 10 User (57,460 points)

The series doesn't look right. The nth term would be n.3^2 not n.3^n. Alternatively, the series would be:

(1)(3)+(2)(3^2)+(3)(3^3)+...+(n)(3^n).

Let's assume it's the first alternative: 3^2(1+2+3+...+n)=9n(n+1)/2. Since this doesn't give us the format expressed in the equation, we have to assume the second alternative:

∑(n)(3^n) for numbers 1, 2, 3, ..., n.

S[n+1]=S[n]+(n+1)3^(n+1), where S[n] means the sum to n terms.

Let's suppose S[n]=(3/4)((2n-1)3^n+1), then:

S[n+1]=(3/4)((2n-1)3^n+1)+(n+1)3^(n+1)=

(1/4)((2n-1)3^(n+1)+3+4(n+1)3^(n+1)=(1/4)((6n+3)3^(n+1)+3)=

(3/4)((2n+1)3^(n+1)+1)=(3/4)((2(n+1)-1)3^(n+1)+1)=S[n+1] according to the supposed formula.

Also, the base case for n=1, S1=(3/4)(3+1)=3, the first term (1)(3).

So the formula works for consecutive values of n, and it's correct for S1, therefore, by induction, the formula is correct for the general sum up to n.

answered Dec 22, 2016 by Rod Top Rated User (486,860 points)
reshown Dec 22, 2016 by Rod
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