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Question
Using the Mathematical induction, show that for any natural number n,
`1/(2.5) + 1/(5.8) + 1/(8.11) + ... + 1/((3"n" - 1)(3"n" + 2)) = "n"/(6"n" + 4)`
Solution
Let P(n) = `1/(2.5) + 1/(5.8) + 1/(8.11) + ... + 1/((3"n" - 1)(3"n" + 2)) = "n"/(6"n" + 4)`
For n = 1
P(1) = `1/((3 xx 1 - 1)(3 xx 1 + 2))`
= `1/((6 xx 1 + 4))`
⇒ `1/(2 xx 5) = 1/10`
⇒ `1/10 = 1/10`
∴ P(1) is true
Let P(n) be true for n = k
∴ P(k) = `1/(2.5) + 1/(5.8) + 1/(8.11) + ... + 1/((3"k" - 1)(3"k" + 2))`
= `"k"/(6"k" + 4)` ......(i)
For n = k + 1
P(k + 1) = `1/(2.5) + 1/(5.8) + 1/(8.11) + ... + 1/((3"k" - 1)(3"k" + 2)) + 1/([(3"k" - 1) - 1][3("k" + 1) + 2])`
= `("k" + 1)/(6"k" + 10)`
= `"k"/(6"k" + 4) + 1/((3"kk" + 2)(3"k" + 5))`
= `1/((3"k" + 2))["k"/2 + 1/(3"k" + 5)]`
= `1/((3"k" + 2))[(3"k"^2 + 5"k" + 2)/(2(3"k" + 5))]`
= `1/((3"k" + 2)) [(3"k"^2 + 3"k" + 2"k" + 2)/(2(3"k" + 5))]`
= `1/((3"l" + 2))[(3"k"("k" + 1) + 2("k" + 1))/(2(3"k" + 5))]`
= `1/((3"k" + 2))[(("k" + 1)(3"k" + 2))/(2(3"k" + 5))]`
= `("k" + 1)/(6"k" + 10)`
∴ P(k + 1) is true
Thus P(k) is true
⇒ P(k + 1) is true.
Hence by principle of mathematical induction,
P(n) is true for all n ∈ N.
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