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Question
Using the Mathematical induction, show that for any natural number n ≥ 2,
`1/(1 + 2) + 1/(1 + 2 + 3) + 1/(1 +2 + 3 + 4) + .... + 1/(1 + 2 + 3 + ... + "n") = ("n" - 1)/("n" + 1)`
Solution
Let P(n) is the statement
`1/(1 + 2) + 1/(1 + 2 + 3) + 1/(1 + 2 + 3 + 4) + ... + 1/(1 + 2 + 3 + ... + "n") = ("n" - 1)/("n" + 1)`
Given n ≥ 2
L.H.S ⇒ P(2) = `1/(1 + 2) = 1/3`
R.H.S ⇒ P(2) = `(2 - 1)/(2 +1) = 1/3`
L.H.S = R.H.S ⇒ P(n) is true for n = 2
Assume that the given tatement is true for n = k
(i.e) `1/(1 + 2) + 1/(1 +2 + 3) + ... + 1/(1 + 2 + 3 + ... + "k") = ("k" - 1)/("k" + 1)` is true
To prove P(k + 1) is true
P(k + 1) = `"P"("k") + ("t"_("k" + 1))`
= `("k" - 1)/("k" + 1) + 1/(1 + 2 + ... + "k" + 1)`
= `("k" - 1)/("k" + 1) + 1/((("k" + 1)("k" + 2))/2)`
= `("k" - 1)/("k" + 1) + 2/(("k" + 1)("k" + 2))`
= `(("k" - 1)("k" + 2) + 2)/(("k" + 1)("k" + 2))`
= `("k"^2 - "k" + 2"k" - 2 + 2)/(("k" + 1)("k" + 2))`
= `("k"^2 + "k")/(("k" + 1)("k" + 2))`
= `("k"("k" + 1))/(("k" + 1)("k" + 2))`
= `"k"/("k" + 1)^2`
= `("k" + 1 - 1)/("k" + 1 + 1)`
⇒ P(k + 1) is true when P(k) is true so by the principle of mathematical induction P(n) is true for n ≥ 2.
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