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The sum of cubes of the first n numbers of Natural Numbers
etL Sn be the sum of cubes of the first n numbers of natural numbers.

That is, Sn 13  23  33 n^3

We know that, (r 1 )^2 (r 1 )^2 (r^2  2 r 1 )(r^2  2 r 1 ) 4 r.

or, (r 1 )^2 r^2 r^2 (r 1 )^2 4 r.r^2 4 r^3 [Multiplying both the sides by r^2 ]

In the above identity, putting r 1 , 2 , 3 ,,n

We get,

2 2 2 2 3

2 2 2 2 3

2 2 2 2 3

2 2 2 2 3

( 1 ) ( 1 ) 4

4. 3 3. 3 4. 3

3. 2 2. 1 4. 2

2. 1 1. 0 4. 1

n nn n n







   

   

Adding, we get, (n 1 )^2 .n^2  12. 02 4 ( 13  23  33 n^3 )

or, (n 1 )^2 .n^2 4 Sn

or,

4

(^2) (  1 ) 2
n n
Sn
?
2
2
( 1 )
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½
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nn
Sn
Necessary formulae :
N.. :B 13  23  33 n^3 ( 1  2  3 n)^2.
Activity : 1. Find the sum of natural even numbers of the first n-numbers.

2. Find the sum of squares of natural odd numbers of the first n-numbers.

2

( 1 )

1 2 3



   

nn

 n

6

( 1 )( 2 1 )

12 22 32 2

 

   

nn n

 n

2

3 3 3 3

2

( 1 )

1 2 3

¿

¾

½

̄

®

­ 

   

n

n

n