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प्रश्न
By the principle of Mathematical induction, prove that, for n ≥ 1
`1^2 + 2^2 + 3^2 + ... + "n"^2 > "n"^2/3`
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उत्तर
Let P(n) is the statement `1^2 + 2^2 + 3^2 + ... + "n"^2 > "n"^2/3`
To prove P(1) is true
P(1) = 12
= `1> 1^3/3 (= 1/3)`
`1 > 1/3` which is true
So P(1) is true
Assume that the given statement is true for n = k
(i.e.) `1^2 + 2^2 + ... + "k"^2 > "k"^3/3` is true
To prove P(k + 1) is true
P(k + 1) = P(k) + (k + 1)
P(k + 1) = `1^2 + 2^2 + ... + "k"^2 + ("k" + 1)^2 > "k"^2/3 + ("k" + 1)^2`
R.H.S = `("k"^3 + 3("k" + 1)^2)/3`
= `("k"^3 + 3("k"^2 + 2"k" + 1))/3`
= `("k"^3 + 3"k"^2 + 6"k" + 3)/3`
= `("k"^3 + 3"k"^2 + 3"k" + 3"k" + 1 + 2)/3`
= `("k"^3 + 3"k"^2 + 6"k" + 3)/3 + (3"k" + 2)/3`
= `("k" + 1)^3/3 + (3"k" + 2)/3 > ("k" + 1)^3/3`
⇒ P(k + 1) = `1^2 + 2^2 + ... + ("k" + 1)^2 > ("k" + 1)^2/3`
⇒ P(k + 1) is true whenever P(k) is true.
So by the principle of mathematical inductions
P(n) is true.
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