Question

Prove that :
1^2 + 2^2 + 3^2 +....n^2 > n^3/3
for all n belongs to N.
​

Answer 1

★ Concept

The above inequality can be proved by several methods like Binomial Theorem, Number Theory and even by advance algebra but here we are going to solve this using the concept of Principle of Mathematical Induction for proving inequalities. In this method first of all we'll prove the inequality true for n = 1 which will come under basic step. Basic step will be then followed by Induction step which will be solved under two steps. In first step we will assume the inequality true for n = K (any natural number) and in second step we will prove that the inequality holds true for n = K + 1 using the values from step 1.

Let's proceed !!

$\rule{190pt}{1pt}$

Let T(n) be the mathematical statement given by

T(n) = 1² + 2² + 3² + ... + n² > n³/3

Step 1 : Basic Step

Substitute n = 1

T(1) : 1² > 1³/3 ➔ 1 > 1/3

∴ T(1) is true.

Step 2 : Induction Step

Let T(n) be true for n = K. Then,

T(n) : 1² + 2² + 3² + ... + k² > k³/3

we shall now prove that T(k+1) is true i.e,

1² + 2² + 3² + ... + k²+ (k+1)² > (k+1)³/3

Now, T(n) is true

⇒ 1² + 2² + 3² + ... + k² > k³/3

let , 1² + 2² + 3² + ... + k² + (k+1)² = P

⇒ P > k³/3 +(k+1)²

⇒ P > 1/3(k³+3k²+6k+3)

⇒ P > 1/3{(k+1)³+(3k+2)}

⇒ 1/3{(k+1)³+(3k+2) > (k+1)³}/3

Therefore, we have proved that

1² + 2² + 3² + ... + k² + (k+1)² > (k+1)³/3

⇒ T(K+1) is true.

Thus, T(K) is true.

Hence, by the principle of mathematical Induction, T(n) is true for all n∈N.

$\underline{\rule{190pt}{2pt}}$

More to know

We can use the concept of principle of mathematical Induction for three purposes :-

• Summation of Series
• Proving Inequalities
• Proving Divisibility.

☞ Principle of Mathematical Induction

A statement T(n) is true for all n∈N, where N is the set of natural numbers, provided :-

❶ T(1) is true

❷ T(k) is true ⇒ T(K+1) is true.