Question

Using mathematical induction, prove that 1² + (1² + 2²) + (1² + 2² + 3²) + ... upto n terms = $\frac{n(n + 1)^{2}(n + 2)}{12}$ for all n ∈ N

Answer 1

Let P(n): 12+22+32+..... +n2>n3312+22+32+..... +n2>n33 be the given mathematical statement.

Step 1:  We need to show that the given mathematical statement holds true for n=1.

LHS: 12=112=1
RHS: 133=13133=13

Since LHS > RHS

The given mathematical statement P(n) holds true for n=1.

Step 2: We assume that the given mathematical statement holds true for n=k.

So, 12+22+32 ..... +k2>k3312+22+32 ..... +k2>k33 

or  12+22+32 ..... +k2=d+k33 for d>012+22+32 ..... +k2=d+k33 for d>0-------(A)

Step 3: We need to prove that the given mathematical statement holds true for n=k+1.

To prove:   12+22+32+ ..... +(k+1)2>(k+1)3312+22+32+ ..... +(k+1)2>(k+1)33

LHS:   12+22+32+ ..... +k2+(k+1)212+22+32+ ..... +k2+(k+1)2

= d+k33+(k+1)2d+k33+(k+1)2  [Using eq. (A)]

= d+k3+3(k+1)23d+k3+3(k+1)23

= d+k3+3(k2+1+2k)3d+k3+3(k2+1+2k)3

= d+k3+3k2+3+6k)3d+k3+3k2+3+6k)3

= d+(k3+3k(k+1)+1)+3k+23d+(k3+3k(k+1)+1)+3k+23

=d+(k+1)3+3k+23d+(k+1)3+3k+23

= d+(k+1)33+k+23d+(k+1)33+k+23

= (k+1)33+(d+k+23)(k+1)33+(d+k+23)

= (k+1)33+D(k+1)33+D   where D =d+k+23d+k+23 and D>0


12+22+32+ ..... +k2+(k+1)212+22+32+ ..... +k2+(k+1)2 = (k+1)33+D(k+1)33+D   where D =d+k+23d+k+23 and D>0

or 12+22+32+ ..... +k2+(k+1)212+22+32+ ..... +k2+(k+1)2 > (k+1)33(k+1)33

Thus,P(n) is true for n=k+1

Hence proved!

Answer 2

Solution :

1²+(1²+2²)+..+up to n terms

= [n(n+1)²(n+2)]/12

=> 1²+(1²+2²)+..+(1²+2²+..+(1²+2²+..+n²)

= [n(n+1)²(n+2)/12]

=> 1²+(1²+2²)+...+[n(n+1)(2n+1)/6]

= [n(n+1)²(n+2)/12]

Let S(k) be the given statement

For n = 1

LHS = 1 ,

RHS = [ 1(1+1)²(1+2)/12 ]

= ( 1×4×3)/12

= 1

LHS = RHS

S(1) is true.

Assume that S(k) is true .

1²+(1²+2²)+...+[(k+1)(k+2)(2k+3)/6]

= [k(k+1)²(k+2)/12]

Adding ( k+1)th term i.e

= [(k+1)(k+2)(2k+3)/6] on both sides

1²+(1²+2²)+..+[(k+1)(k+2)(2k+3)/6]

= [k(k+1)²(k+2)/12]+[(k+1)(k+2)(2k+3)/6]

= [k(k+1)²(k+2)+2(k+1)(k+2)(2k+3)/12]

=[(k+1)(k+2)(k²+k+4k+6)/12]

=[(k+1)(k+2)(k²+5k+6)/12]

= [(k+1)(k+2)(k+2)(k+3)/12]

= [(k+1)(k+2)²(k+3)]/12

Therefore ,

The formula holds for n = k+1.

By the principal of mathematical

induction,

S(n) is true for all n € N

i.e the formula

1²+(1²+2²)+..+up to n terms

= [n(n+1)²(n+2)/12] is true for all n€N

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