1. using PMI prove that: 1.3+3.5+5.7+...+(2n-1)(2n+1)=n(4n^2+6n-1)/3?

Question

Ravi · Answer

I assume you must have gotten through the step stating that it is true for n=k
And we have to obtain
(k+1)(4(k+1)^2+6(k+1)-1)/3 i.e. it is true for n=k+1

Then extend the series till (k+1)th term
Till n=k term, sum has been obtained, use it and rewrite the series as
k(4k^2+6k-1)/3 + .(2(k+1)-1)(2(k+1)+1)

Solving this further will obtain a cubic eqn which has to be factorised by Trial and error
[Hint: Look up that we have to prove that Sum till n=k+1 is (k+1)(4(k+1)^2+6(k+1)-1)/3 ]

So one factor can be (k+1). Divide by this to fatorise the cubic eqn. and reach your answer.

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1. using PMI prove that: 1.3+3.5+5.7+...+(2n-1)(2n+1)=n(4n^2+6n-1)/3?

1. using PMI prove that:

1.3+3.5+5.7+...+(2n-1)(2n+1)=n(4n^2+6n-1)/3?

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