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Exercises · Q5

Q.Prove the following by using the principle of mathematical induction for all n∈Nn \in N: 1.3+2.32+3.33+…+n.3n=(2n−1)3n+1+341.3 + 2.3^2 + 3.3^3 + \ldots + n.3^n = \dfrac{(2n-1)3^{n+1}+3}{4}

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Let P(n)P(n) be the statement

1⋅3+2⋅32+3⋅33+…+n⋅3n=(2n−1)3n+1+34.1\cdot3+2\cdot3^2+3\cdot3^3+\ldots+n\cdot3^n=\frac{(2n-1)3^{n+1}+3}{4}.

Base case: For n=1n=1,

LHS=1⋅3=3,RHS=(2⋅1−1)32+34=1⋅9+34=124=3.\text{LHS}=1\cdot3=3,\qquad \text{RHS}=\frac{(2\cdot1-1)3^{2}+3}{4}=\frac{1\cdot9+3}{4}=\frac{12}{4}=3.

So P(1)P(1) is true.

Inductive step: Assume P(k)P(k) is true for some k≥1k\ge1:

1⋅3+2⋅32+…+k⋅3k=(2k−1)3k+1+34.(Induction Hypothesis)1\cdot3+2\cdot3^2+\ldots+k\cdot3^k=\frac{(2k-1)3^{k+1}+3}{4}. \qquad \text{(Induction Hypothesis)}

We must show the sum up to (k+1)(k+1) equals (2k+1)3k+2+34\dfrac{(2k+1)3^{k+2}+3}{4}.

Adding the next term (k+1)⋅3k+1(k+1)\cdot3^{k+1} to both sides of the hypothesis:

(2k−1)3k+1+34+(k+1)3k+1=(2k−1)3k+1+3+4(k+1)3k+14\frac{(2k-1)3^{k+1}+3}{4}+(k+1)3^{k+1}=\frac{(2k-1)3^{k+1}+3+4(k+1)3^{k+1}}{4}

=3k+1[(2k−1)+4(k+1)]+34=\frac{3^{k+1}\big[(2k-1)+4(k+1)\big]+3}{4}

Simplify the bracket: …

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