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Prove by the method of induction, for all n ∈ N.

3 + 7 + 11 + ……… to n terms = n(2n + 1)

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Let P(n) = 3 + 7 + 11 + ……… to n terms =

n(2n +1), for all n ∈ N.

But 3, 7, 11, …. are in A.P.

∴ a = 3 and d = 4

Let tn be the nth term.

∴ t = a + (n – 1)d = 3 + (n – 1)4 = 4n – 1 ∴ P(n) = 3 + 7 + 11 + ……. + (4n – 1) = n(2n + 1)

Step I:

Put n = 1

L.H.S. = 3

R.H.S. = 1[2(1)+ 1] = 3

∴ L.H.S. = R.H.S.

∴ P(n) is true for n = 1.

Step II: Let us assume that P(n) is true for n = k.

∴ 3 + 7 + 11 + ….. + (4k – 1) = k(2k + 1) …..(i)

Step III: We have to prove that P(n) is true for n = k + 1,

i.e., to prove that

3 + 7 + 11 + …+ [4(k + 1) – 1] = (k + 1)(2k + 3)

L.H.S. = 3 + 7 + 11 + …… + [4(k + 1) – 1]

= 3 + 7 + 11 + ….. + (4k – 1) + [4(k+ 1) – 1]

= k(2k + 1) + (4k + 4 – 1) …..[From (i)]

= 2k2 + k + 4k + 3

= 2k2 + 2k + 3k + 3

= 2k(k + 1) + 3(k + 1)

= (k + 1) (2k + 3)

= R.H.S.

∴ P(n) is true for n = k + 1.

Step IV: From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.

∴ 3 + 7 + 11 + ….. to n terms = n(2n + 1) for all n ∈ N.

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