How do you find the number of terms n given the sum #s_n=2178# of the series #2+16+30+44+58+...#?

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Answer 1 · 2016-07-29T14:45:29

there is none

First lets find the equation from the series. Immediately we see a difference of #14# per number starting at #2# so this looks something like this

#sum_1^n14(n-1)+2#

now we need to solve for #n# given

#sum_1^n14(n-1)+2 = 2178#

#(14sum_1^n(n-1))+2n = 2178#
#(14sum_1^n n)-14n+2n = 2178#
#14(n(n+1))/2-12n = 2178#
#7n^2-5n - 2178 = 0#

using the quadratic formula #n= -b +- sqrt((b^2 -4ac)/(2a))#.

#n= -5 +- sqrt((25 +4*7*2178)/(2))#

#n~~174.65+-5#

so the summation is off because the closes number to
#2178# is the first #179# terms but this is equal to #2148#