solve 1+2*2+3*3*3+4*4*4*4+....... n terms?

Answer 1

I'm keeping an eye on this. I certainly haven't cracked it, and so far neither has anyone else.

Answer 2

The only way I can see to mathematically express this is via a summation:

the n-th term is given by n^n, and to get the sum of the sequence, you need to add to that the sum of (n-x)^(n-x) for 0 < x <= n

combining the two into a single statement:

y(n) = SUM(0->x->n) (n-x)^(n-x)

example for n = 4:

y(4) = (4-0)^(4-0) + (4-1)^(4-1) + (4-2)^(4-2) + (4-3)^(4-3) + (4-4)^(4-4)
y(4) = 4^4 + 3^3 + 2^2 + 1^1
y(4) = 16 + 9 + 4 + 1
y(4) = 30

Answer 3

The formula is :

n(n+1)(2n+1)/6

E.g., if n = 4, the answer is 4*5*9/6 = 180/6 = 30 = (1+4+9+16)

Answer 4

term tn= n^n

sum = n^ * [1+ (n-1)^n-1 * n^-n + (n-2)^(n-2)*n^-n)]

Answer 5

If you want people to do your homework, at least try to state the problem cleanly.

Answer 6

(n^0)^(n^0) +[(n^0 + n^0)]^(n^0 + n^0) + [(n^0 + n^0 + n^0)]^(n^0 + n^0 + n^0) + [(n^0 +n^0 + n^0 +n^0)]^(n^0 + n^0 + n^0 + n^0) + ... n

where n is any natural number.

Answer 7

sumation would be the way to do it out... rest i could nt solve

Answer 8

first you want to do your multiplication mmmkay. then you add everything together got it tough guy.

Answer 9

=2560+

Answer 10

i have solved it out!!