consecutive square numbers
16 + 9 + 4 + 1
The simplest of infinitely many possible solutions, is y = x2
1 .... - 1, 1+4(square of 2) - 5, 1+4+9(square of 3) - 14, 1+4+9+16(square of 4) - 30, 1+4+9+16+25(sqaure of 5)- 55,
the next number in the sequence is 25............starting at the beginning, you add the odd numbers to the existing number to get the answer. Add "1" to 0, get 1; add "3" to 1, get 4; add "5" to 4, get 9; add "7" to 9 to get 16........and add "9" to 16 to get 25.........
consecutive square numbers
add 3,5,7,9and then to 25 add 11, adding odd numbers onto the last . 1+3=4, 4+5=9, 9+7=16, 16+9=25, 25+11=36 and so on
t(n) = n2
9/16 + 1/4 = 9/16 + 4/16 = 13/16.
t(n) = (n+1)2 where n = 1, 2, 3, ...
16 + 9 + 4 + 1
The factors of 9 are 1, 3, 9 The factors of 16 are 1, 2, 4, 8, 16 The common factor of 9 and 16 is 1
four 1 + 2 + 4 + 9 = 16 four numbers so divide 16 by 4 16/4 = 4 average
Its square factors are: 1, 4, 9, 16 and 144
9 1/16. (1 times 5 is 5, and 13/16 times 5 is 65/16 which is 4 1/16; the 4 adds to the 5 to give 9, and then the 1/16 makes it 9 1/16.)
The simplest of infinitely many possible solutions, is y = x2
9 = 1 x 3 x 3 16 = 1 x 4 x 4 The only common factor of 9 and 16 is 1.