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Exercises · Q8

Q.Prove the following by using the principle of mathematical induction for all n∈Nn \in N: 1.2+2.22+3.23+…+n.2n=(n−1)2n+1+21.2 + 2.2^2 + 3.2^3 + \ldots + n.2^n = (n-1)2^{n+1} + 2

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

1⋅2+2⋅22+3⋅23+…+n⋅2n=(n−1)2n+1+2.1\cdot2+2\cdot2^2+3\cdot2^3+\ldots+n\cdot2^n=(n-1)2^{n+1}+2.

Base case: For n=1n=1,

LHS=1⋅2=2,RHS=(1−1)22+2=0+2=2.\text{LHS}=1\cdot2=2,\qquad \text{RHS}=(1-1)2^{2}+2=0+2=2.

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

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

1⋅2+…+k⋅2k=(k−1)2k+1+2.(Induction Hypothesis)1\cdot2+\ldots+k\cdot2^k=(k-1)2^{k+1}+2. \qquad \text{(Induction Hypothesis)}

We must show the sum up to (k+1)(k+1) equals k⋅2k+2+2k\cdot2^{k+2}+2.

Adding the next term (k+1)⋅2k+1(k+1)\cdot2^{k+1} to both sides of the hypothesis:

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

=2k+1⋅2k+2=k⋅2k+2+2.=2^{k+1}\cdot2k+2=k\cdot2^{k+2}+2. …

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