Given an arithmetic sequence
#a, a+d, a+2d, ...#
where #a# is the initial term and #d# is the difference between terms,
the sum of the first #n# terms of the sequence is given by
#a + (a+d) + ... + a+(n-1)d = sum_(k=0)^(n-1)(a+kd) = n((a_1+a_n)/2)#
where #a_i# is the #i#th term in the sequence
(so #a_1 = a# and #a_n = a + (n-1)d#)
(a derivation of this formula is given below)
Applying the formula here with #a = 3# and #d = 6# we get
#3 + 9 + 15 + ... + (3 + (22-1)6) = sum_(k=0)^(21)(3+6k)#
#= 22((3 + (3+21*6))/2) = 1452#
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The following is a derivation of the arithmetic sum formula, and is not strictly necessary for the solution of the above problem.
#sum_(k=0)^(n-1)(a+kd) = sum_(k=0)^(n-1)a + sum_(k=0)^(n-1)kd = na + dsum_(k=0)^(n-1)k#
The sum of the first #n# integers has the closed form
#sum_(k=1)^nk = (n(n+1))/2#
So applying this to the sum of the first #n-1# integers gives
#sum_(k=0)^(n-1)k = sum_(k=1)^(n-1)k = ((n-1)((n-1)+1))/2 = (n(n-1))/2#
Thus
#sum_(k=0)^(n-1)(a+kd) = na + d(n(n-1))/2 = n((2a + d(n-1))/2)#
#=n((a + (a + (n-1)d))/2) = n((a_1+a_n)/2)#
