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Question
Using the Mathematical induction, show that for any natural number n,
`1/(1*2*3) + 1/(2*3*4) + 1/(3*4*5) + ... + 1/("n"("n" + 1)*("n" + 2)) = ("n"("n" + 3))/(4("n" + 1)("n" + 2))`
Solution
Let P(n) = `1/(1*2*3) + 1/(2*3*4) + 1/(3*4*5) + ... + 1/("n"("n" + 1)*("n" + 2)) = ("n"("n" + 3))/(4("n" + 1)("n" + 2))`
For n = 1
P(1) = `1/(1(1 + 1)(1 + 2))`
= `(1(1 + 3))/(4(1 + 1)(1 + 2))`
⇒ `1/(1 xx 2 xx 3) = 4/(4 xx 2 xx 3)`
⇒ `1/6 = 1/6`
∴ P(1) is true
Let P(n) be true for n = k
∴ P(k) = `1/(1*2*3) + 1/(2*3*4) + 1/(3*4*5) + ... + 1/("k"("k" + 1)("k" + 2)) + ("k"("k" + 3))/(4("k" + 1)("k" + 2))` .....(1)
For n = k + 1
R.H.S = `(("k" + 1)("k" + 4))/(4("k" + 2)("k" + 3))`
L.H.S = `("k"("k" + 3))/(4("k" + 1)("k" + 3)) + 1/(("k" + 1)("k" + 2)("k" + 3))` .....[Using (i)]
= `1/(("k" + 1)("k" + 2)) [("k"^3 + 3"k")/4 + 1/("k" + 3)]`
= `1/(("k" + 1)("k" + 2))[("k"^2 + 6"k"^2 + 9"k" + 4)/(4("k" + 3))]`
= `1/(("k" + 1)("k" + 2)) [(("k" + 1)^2 ("k" + 4))/(4("k" + 3))]`
= `[(("k" + 1)("k" + 4))/(4("k" + 2)("k" + 3))]`
∴ 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 ∈ z
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