IIT JEE , AIEEE Forum - Mathematics , Algebra

namey

use a suitable identity to prove thet

te sum of the cubes of the first 'n' natural numbers is

(

n)²

nadeemoidu

n³ = n²(n+1)²/4

n= n(n+1)/2

therefore

n³= (

n)²

prathyu

nadeemoidu is correct.

goiit_user

I've used 2 formulae for the proof,

n = n(n+1)/2 ...which can be derived easily using Sum of AP

n²₌n(n+1)(2n+1)/6         ...which can be derived using the method i've used to solve                                          ur question

Now,

(1+x)⁴ -x⁴     =   (x⁴ + 4x³+ 6x²+4x+1) - x⁴
(1+x)⁴ -x⁴     =   4x³+ 6x²+4x+1

Putting x =1,2,......,n

2⁴ - 1⁴   = 4.1³ + 6.1² + 4.1 +1
3⁴ - 2⁴   = 4.2³ + 6.2² + 4.2 +1
.
.
.
.
.
(1+n)⁴ - n⁴ = 4.n³ +6.n² + 4.n +1

Adding all these we see the diagonal terms on LHS get cancelled, giving only two terms

(1+n)⁴ - 1⁴ = 4 (1³ + 2³ + ..... n³) + 6.(1² + 2² +.....n²) + 4(1 + 2+ .....n ) +                             (1+1+.....n terms)

             Let

n³ = 1³ + 2³ + ..... n³ = k

n⁴ + 4n³ + 6n² +4n = 4k + 6. (n)(n+1)(2n+1)/6 + 4 (n)(n+1)/2 + n
                       = 4k + n(n+1)(2n+1) + 2n(n+1) + n
                           = 4k + (2n³ + 3n² + n) + (2n² + 2n) + n
                           = 4k + 2n³ +5n² + 4n
4k = n⁴ + 2n³ + n²
4k = n²(n² + 2n + 1)
4k = n²(n+1)²
k   = (n²(n+1)²)/4
k   = [n(n+1)/2]²

n³=(

n)²

Thus, proved

Plz rate me!!

edison

First we prove the formula for the sum of cubes of first n natural numbers as below

To Prove this formula for the sum of consecutive cubes:

=

n²(n + 1)²
4

Proof. Assume that the statement -- the formula -- is true for n = k. That is, assume that the sum up to k -- S(k) -- is true:

S(k)	=	k²(k + 1)² 4	(1)

Then we must show that it is also true for n = k + 1:

S(k + 1)	=	(k + 1)²(k + 2)² 4	(2)

To do that, add the next cube to S(k), line (1):

S(k + 1)	=	S(k) + (k + 1)³

	=	k²(k + 1)² 4	+ (k + 1)³

	=	k²(k + 1)² + 4(k + 1)³ 4

	=	(k + 1)²[k² + 4(k + 1)] 4

-- on having factored (k + 1)²,

	=	(k + 1)²(k² + 4k + 4) 4

	=	(k + 1)²(k + 2)² 4

This is line (2), which is what we wanted to show.

Next, show that the formula is true for n = 1.

S(1)	=	1³	=	1²· 2² 4

		1	=	1· 4 4

-- which is true. The formula, therefore, is true for every natural number.

Now

To Prove that the sum of the first n natural numbers is given by this formula:

S(n) = 1 + 2 + 3 + . . . + n

=

n(n + 1)
2

We will call this statement S(n), because it depends on n.

Proof. We will do Steps 1) and 2) above. First, we will assume that the statement is true for n = k; that is, we will assume that S(k) is true:

S(k) = 1 + 2 + 3 + . . . + k

=

k(k + 1)
2

. . . . . (1)

This is the induction assumption. Assuming this, we must prove that S(k + 1) is also true. That is, we must show:

S(k + 1) = 1 + 2 + 3 + . . . + (k + 1)

=

(k + 1)(k + 2)
2

. . . (2)

To do that, we will simply add the next term (k + 1) to both sides of the induction assumption, line (1):

This is line (2), which is the first thing we wanted to show.

Next, we must show that the statement is true for n = 1. We have

S(1) = 1 = ½· 1· 2

The formula therefore is true for n = 1. We have now fulfilled both conditions of the principle of mathematical induction. S(n) is therefore true for every natural number.

Thus from above to proofs we find that

n³ = n²(n+1)²/4 = (

n)²

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