Find 1^2 + 2^2 + 3^3 + 4^2 +5^2 + ... 97^2 + 98^2.

I only know how to use the sum/seq functions on my graphing calculator to solve this. Could someone show me how to solve this without that cheat.

THANKS

Printable View

- February 3rd 2007, 06:15 PMMr_Greensum of sequence
Find 1^2 + 2^2 + 3^3 + 4^2 +5^2 + ... 97^2 + 98^2.

I only know how to use the sum/seq functions on my graphing calculator to solve this. Could someone show me how to solve this without that cheat.

THANKS - February 3rd 2007, 06:20 PMThePerfectHacker
- February 4th 2007, 01:31 PMtopsquark
Or you can always use the "old fashioned way." Simple addition :eek: :

1 + 4 + 9 + 16 + ... + 9409 + 9604

but I wouldn't recommend it! :)

-Dan - February 4th 2007, 02:56 PMSoroban
Hello, Mr_Green!

Quote:

Find

ThePerfectHacker gave you the formula for this problem.

If you're not familiar with it, you could derive it,

. . but it's faster to simply add the ninety-eight squares!

For anyone who is interested, here is one method.

List the first few partial sums.

Take differences of consecutive pairs of the numbers.

Take differences of the differences, and so on

. . until you reach a row of constant differences.

. .

. . .

. . . . .

. . . . . .

The constants appeared in the__third__differences.

Hence, the generating function is of the__third__degree (a cubic).

The general form of a cubic is: .

We know the first four sums of the sequence; plug them in.

Therefore, the function is: .