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Question
Prove by method of induction, for all n ∈ N:
1.3 + 3.5 + 5.7 + ..... to n terms = `"n"/3(4"n"^2 + 6"n" - 1)`
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Solution
Let P(n) ≡ 1.3 + 3.5 + 5.7 + ..... to n terms = `"n"/3(4"n"^2 + 6"n" - 1)`, for all n ∈ N
But the first factor in each term
i.e., 1, 3, 5 … are in A.P. with a = 1 and d = 2.
∴ nth term = a + (n –1)d = 1 +(n – 1)2 = (2n – 1)
Also second factor in each term
i.e., 3, 5, 7, … are in A.P. with a = 3 and d = 2.
∴ nth term = a + (n – 1)d = 3 + (n – 1)2 = (2n+1)
∴ nth term, tn = (2n – 1) (2n + 1)
∴ P(n) ≡ 1.3 + 3.5 + 5.7 + .... + (2n – 1) (2n + 1) = `"n"/3(4"n"^2 + 6"n" - 1)`
Step I:
Put n = 1
L.H.S. = 1.3 = 3
R.H.S. = `1/3[4(1)^2 + 6(1) - 1]` = 3 = L.H.S.
∴ P(n) is true for n = 1.
Step II:
Let us consider that P(n) is true for n = k
∴ 1.3 + 3.5 + 5.7 + ..... + (2k – 1)(2k + 1)
= `"k"/3(4"k"^2 + 6"k" - 1)` ...(i)
Step III:
We have to prove that P(n) is true for n = k + 1
i.e., to prove that
1.3 + 3.5 + 5.7 + …. + [2(k + 1) – 1][2(k + 1) + 1]
= `(("k" + 1))/3[4("k" + 1)^2 + 6("k" + 1) - 1]`
= `(("k" + 1))/3(4"k"^2 + 8"k" + 4 + 6"k" + 6 - 1)`
= `(("k" + 1))/3(4"k"^2 + 14"k" + 9)`
L.H.S. = 1.3 + 3.5 + 5.7 + ... + [2(k + 1) – 1][2(k + 1) + 1]
= 1.3 + 3.5 + 5.7 + ... + (2k – 1)(2k + 1) + (2k + 1)(2k + 3)
= `"k"/3(4"k"^2 + 6"k" - 1) + (2"k" + 1) (2"k" + 3)` ...[From (i)]
= `1/3[4"k"^3 + 6"k"^2 - "k" + 3(2"k" + 1)(2"k" + 3)]`
= `1/3(4"k"^3 + 6"k"^2 - "k" + 12"k"^2 + 24"k" + 9)`
= `1/3(4"k"^3 + 18"k"^2 + 23"k" + 9)`
= `1/3("k" + 1)(4"k"^2 + 14"k" + 9)`
= R.H.S.
∴ P(n) is true for n = k + 1
Step IV:
From all steps above by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ 1.3 + 3.5 + 5.7 + ..... to n terms = `"n"/3(4"n"^2 + 6"n" - 1)` for all n ∈ N.
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