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Question
Answer the following:
Prove, by method of induction, for all n ∈ N
8 + 17 + 26 + … + (9n – 1) = `"n"/2(9"n" + 7)`
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Solution
Let P(n) ≡ 8 + 17 + 26 + … + (9n – 1) = `"n"/2(9"n" + 7)`, for all n ∈ N
Step I:
Put n = 1
L.H.S. = 8
R.H.S. = `1/2[9(1)+ 7]` = 8 = L.H.S.
∴ P(n) is true for n = 1
Step II:
Let us consider that P(n) is true for n = k
∴ 8 + 17 + 26 + … + (9k – 1) = `"k"/2(9"k" + 7)` ...(i)
Step III:
We have to prove that P(n) is true for n = k + 1
i.e., 8 + 17 + 26 + … + [9(k + 1) – 1]
= `(("k" + 1))/2 [9("k" + 1) + 7]`
= `(("k" + 1))/2 (9"k" + 16)`
L.H.S. = 8 + 17 + 26 + … + [9(k + 1) – 1]
= 8 + 17 + 26 + … + (9k – 1) + [9(k + 1) – 1]
= `"k"/2 (9"k" + 7) + (9"k" + 8)` ...[From (i)]
= `(9"k"^2 + 7"k" + 18"k" + 16)/2`
= `(9"k"^2 + 25"k" + 16)/2`
= `(9"k"^2 + 9"k" + 16"k" + 16)/2`
= `(9"k"("k" + 1) + 16("k" + 1))/2`
= `(("k" + 1))/2(9"k" + 16)`
= 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.
∴ 8 + 17 + 26 + … + (9n – 1) = `"n"/2(9"n" + 7)` for all n ∈ N
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