term(1) = 1^2 -2 = -1
term(2) = 2^2 - 2 = 2
term(3) = 3^2 - 2 = 7
term(4) = 4^2 - 2 = 14
...
term(n) = n^2 - 2
62 = 8^2 - 2 , so we have 8 terms
sigma (n^2 - 2) for 1 to n
= n(n+1)(2n+1)/6 - 2n
recalling that sum n^2 = n(n+1)(2n+1)/6
checking for sum(5)
= 5(6)(11)/6 - 10 = 55-10 = 45
-1+2+7+14+23 = 45
write the finite series -1+2+7+14+23+...+62 in summation notation
1 answer