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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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.
- 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
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½
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n
n
n