what is the sum of the cubes of the first n natural numbers
Answers
Answered by
1
[ n(n+1)/2]^2 is the required formula
Answered by
0
hi mate! your answer↓↓
a=1, d=1,Sn=?
Sn = n/2[2a+(n-1) d]
=n/2[2×1+(n-1) 1]
=n/2[2+n-1]
=n/2[1+n]
a=1, d=1,Sn=?
Sn = n/2[2a+(n-1) d]
=n/2[2×1+(n-1) 1]
=n/2[2+n-1]
=n/2[1+n]
priyanshfiitjee:
absolutely wrong
why it is wrong?
yes,mine is right
It is the sum of n natural numbers not cubes sis
i know
the question demanded aum of cubes so mine is right
sorry,i haven't read proper
i thought the sum of n natural numbers only
i didn't read cube
np
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