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Question

Prove by using the principle of mathematical induction nN

2+5+8+11+...+(3n1)=12n(3n+1)

Or

Using principle of mathematical induction, prove that 4n+15n1is divisible by 9 for all natural numbers n.

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Solution

Let p(n) be the statement given by

p(n)2+5+8+11+...+(3n1)=12n(3n+1)

For n=1

LHS:P(1)=2 and RHS:p(1)=12(1)(3(1)+1)=2

LHS=RHS

Thus p(1) is true.

Assume that p(k) is true for some natural number k.

2+5+8+11+...+(3k1)=12k(3k+1) ...(i)

We will now show that p(k+1) is true, whenever p(k) is true.

On adding (3k+2) both sides , of Eq.(i) , we get 2+5+8+11+...+(3k1)+(3k+2)=12k(3k+1)+(3k+2)

2+5+8+11+...(3k1)+(3k+2)=3k2+k+6k+42

=3k2+3k+4k+42=(k+1)(3k+4)2=12(k+1)(3k+4)

Thus p(k+1) is true.

Hence , by principle of mathematical induction p(n) is true nN.

Or

Let p(n) be the statement given by

p(n):4+151=18, which is divisible by 9.

Thus P(1) is true.

Assume that p(k) is true for some natural number k.

p(k):4k+151 is divisible by 9.

4k+151=9mfor some mN ...(i)

We will now show that P(k+1) is true, whenever p(k) is true.

Now 4k+1+15(k+1)1=4k.4+15k+151

=9m15k+1)4+15k+14=36m60k+4+15k+14, [usingEq.(1)]

=9(4m5k+2).which is divisible 9.

p(k+1) is true, whenever p(k) is true.

Hence by principle of mathematical induction. p(n) is true, nN..


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