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Question
Answer the following:
Prove, by method of induction, for all n ∈ N
`1/(3.4.5) + 2/(4.5.6) + 3/(5.6.7) + ... + "n"/(("n" + 2)("n" + 3)("n" + 4)) = ("n"("n" + 1))/(6("n" + 3)("n" + 4))`
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Solution
Let P(n) ≡ `1/(3.4.5) + 2/(4.5.6) + 3/(5.6.7) + ... + "n"/(("n" + 2)("n" + 3)("n" + 4)) = ("n"("n" + 1))/(6("n" + 3)("n" + 4))`, for all n ∈ N.
Step 1:
For n = 1, L.H.S. = `1/(3.4.5) = 1/60`
R.H.S. = `(1(1+1))/(6(1+3)(1+4))=2/(6(4)(5))=1/60`
∴ L.H.S. = R.H.S. for n = 1.
∴ P(1) is true.
Step 2:
Let us assume that for some k ∈ N, P(k) is true,
i.e., `1/(3.4.5) + 2/(4.5.6) + 3/(5.6.7) + ... + "k"/(("k" + 2)("k" + 3)("k" + 4)) = ("k"("k" + 1))/(6("k" + 3)("k" + 4))` ...(1)
Step 3:
To prove that P(k + 1) is true, i.e., to prove that
`1/(3.4.5) + 2/(4.5.6) + 3/(5.6.7) + ... + "k"/(("k" + 2)("k" + 3)("k" + 4)) + ("k" + 1)/(("k" + 3)("k" + 4)("k" + 5)) = (("k" + 1)("k" + 2))/(6("k" + 4)("k" + 5))`
Now, L.H.S. = `1/(3.4.5) + 2/(4.5.6) + 3/(5.6.7) + ... + "k"/(("k" + 2)("k" + 3)("k" + 4)) + ("k" + 1)/(("k" + 3)("k" + 4)("k" + 5))`
= `("k"("k" + 1))/(6("k" + 3)("k" + 4)) + ("k" + 1)/(("k" + 3)("k" + 4)("k" + 5))` ...[By (1)]
= `("k" + 1)/(("k" + 3)("k" + 4))["k"/6 + 1/("k" + 5)]`
= `("k" + 1)/(("k" + 3)("k" + 4))[("k"^2 + 5"k" + 6)/(6("k" + 5))]`
= `("k" + 1)/(("k" + 3)("k" + 4)) xx (("k" + 2)("k" + 3))/(6("k" + 5))`
= `(("k" + 1)("k" + 2))/(6("k" + 4)("k" + 5))`
= R.H.S.
∴ P(k + 1) is true.
Step 4:
From all the above steps and by the principle of mathematical induction P(n) is true for all n ∈ N,
i.e., `1/(3.4.5) + 2/(4.5.6) + 3/(5.6.7) + ... + "n"/(("n" + 2)("n" + 3)("n" + 4)) = ("n"("n" + 1))/(6("n" + 3)("n" + 4))`, for all n ∈ N.
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