How do I find the sum of the series: 4+5+6+8+9+10+12+13+14+⋯+168+169+170. since D is changing from +1, +1 to +2 ?

1 Answer
Sasha P.
Oct 20, 2015

#10962#, see the explanation.

Explanation:

If we make three sums:

#4+8+12+... = sum_(k=1)^n 4k#
#5+9+13+... = sum_(k=1)^n (4k+1) = sum_(k=1)^n 4k + sum_(k=1)^n 1 = sum_(k=1)^n 4k + n#
#6+10+14+... = sum_(k=1)^n (4k+2) = sum_(k=1)^n 4k + sum_(k=1)^n 2 = sum_(k=1)^n 4k + 2n#

Then :

#sum = sum_(k=1)^n 4k + sum_(k=1)^n 4k + n + sum_(k=1)^n 4k + 2n = 3sum_(k=1)^n 4k + 3n#

#sum = 12sum_(k=1)^n k + 3n = 12 (n(n+1))/2 +3n = 6n(n+1)+3n#

#sum = 3n(2(n+1)+1) = 3n(2n+3)#

Note: sum of first #n# integers is #(n(n+1))/2#.

We have to find #n# and it's the number of members in the first sum, so:

#n=168/4 = 42#

Finally:

#sum = 3*42*(2*42+3) = 126*87 = 10962#

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