Thus, T = k(k+1)(2k+1)/6 which is the desired result. The constant c is equal to the n 0 term of the sequence. Thus adding all these terms together gives: Here's a quick summary of what you need to know to get the explicit form of a quadratic sequence: The second difference is equal to 2a. The first term in each line is 3 times the sum of squares (eg what we are trying to work out). The next term in each line is 3 times the sum of integers. Starting from the right-hand side of this equation.
There are only two terms which cannot be cancelled, namely: Now let's sum the two sides vertically.Ĭompare each row with the next and you will notice that the first term in one row is subtracted in the next row (eg most of the terms cancel each other out). Let's start out by looking at the formula for the cubic equation: We have thus arrived at the required result. When we add all these equations together, we obtain: On the other side, there are k lots of 1 and 2 multiplied by the sum (1 + 2 + 3 +. (EG- 19-154, 2nd differences: 4,4)By halving our second difference we know the number that goes before X2 in our quadratic sequence. The only terms that do not cancel from the left hand side are (k+1)^2 and -1^2. Now write the following list, one for each value of n from n=1 to n=k.Īdd both sides of the above list, noticing that many of the terms on the left hand side cancel out. Using a similar method to one worked out by Guass we can then adapt this (and then use this proven result) to go further and find the sum of squares.
To start this proof you will need to find an expression for the sum of all integers.
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