Answer :

Let P(n) = 1 + 2+ 3 +…. + n < ( 2n + 1)

^{2}

P(1) is true

Let us assume p(k) is true.

1 + 2+ 3 +…. + k < ( 2k + 1)

^{2}

To prove P(k+1) is true using P(k)

P(k+1) = 1 + 2+ 3 +…. + k +k+1< ( 2(k + 1)+1)

^{2 }= 1+ (1 + 2+ 3 +…. + k) +k < ( 2k + 3)

^{2}…….. 1

L .H.S 1+ ( 2k + 1)

^{2}+ k

=

^{ }= ( 4k

^{2}+ 1 +4k + 8k + 8)

^{ }= (4k

^{2}+ 12k +9)

^{ }= (2k + 3)

^{2}

Which is the R.H.S of……. 1

P(K+1) is true.

By the Principle of mathematical induction, P(n) is true for all values of n where n N

Hence proved.

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