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2 + 4 + 6 +...+ 2n = n^2 + n for all natural numbers n.

Answers (1)

P(n)=2 + 4 + 6 +...+ 2n = n^2 + n

 

Now, we will substitute different values for n,

P(0) = 0 = 0^2 + 0, is true

P(1) = 2 = 1^2 + 1, is true

P(2) = 2+4 = 2^2 + 2, is true

Now, let us consider,

P(k) = 2 + 4 + 6 + .... + 2k = k^2 + k to be true,

Thus,

P(k+1) is

2 + 4 + 6 + .... + 2k + 2(k+1) = k^2 + k + 2k + 2

                                          = (k^2 + 2k + 1) + (k+1)

                                          = (k + 1)^2 + (k + 1)

Thus, P(k+1) is true if P(k) is true

Hence, by mathematical induction,

For each natural no. n it is true that, P(n)=2 + 4 + 6 +...+ 2n = n^2 + n

 

Posted by

infoexpert21

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