2006-09-24 17:45:49 · 7 answers · asked by Janarthanan S 1 in Science & Mathematics ➔ Mathematics
it is (1+2+3.... +n) ^ 2
or n^2(n+1)^2/4
this is serious and no jokes you can test it using Principle of mathemetical induction
2006-09-24 17:50:17 · answer #1 · answered by Mein Hoon Na 7 · 3⤊ 1⤋
Sum(n^3+(n-1)^3+(n-2)^3+(n-4)^3+. . . 1)
=(n(n+1)/2)^2
2006-09-24 18:01:33 · answer #2 · answered by Helmut 7 · 0⤊ 0⤋
(1+2+3+4+5........+n)^3
2006-09-24 17:58:42 · answer #3 · answered by Brenmore 5 · 0⤊ 0⤋
1^3 + 2^3 + 3^3 + 4^3 + ... + n^3 =
(1 + 2 + 3 + ... + n)^2
which, in turn, is the same as
n^2 * (n+1)^2 / 4
Proof: it is clearly true that 1^3 = 1^2. Now assume that the statement is true for n; we prove it is true for n+1.
Now
1^3 + ... + n^3 + (n+1)^3 =
... = n^2 * (n+1)^2 / 4 + (n+1)^3 [assumption]
... = (n^4 + 2 n^3 + n^2)/4 + (n^3 + 3n^2 + 3n + 1)
... = 1/4 n^4 + 1 1/2 n^3 + 3 1/4 n^2 + 3 n + 1
... = (n^4 + 6 n^3 + 13 n^2 + 12 n + 4)/4
... = (n+1) (n+1) (n+2) (n+2) / 4
QED
2006-09-24 17:59:50 · answer #4 · answered by dutch_prof 4 · 0⤊ 1⤋
1^3+2^3+3^3........+n^3 = (n*(n+1)*((2*n)+1))/6
2006-09-24 23:45:05 · answer #5 · answered by raj 1 · 0⤊ 0⤋
(n(n+1)/2)²
2006-09-24 17:51:06 · answer #6 · answered by Pascal 7 · 0⤊ 0⤋
sum = n(n+1)(2n+1)/6
2006-09-24 23:35:28 · answer #7 · answered by Anonymous · 1⤊ 0⤋