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 ?

Question

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 ?

Emma Aldridge · Accepted Answer

#10962#, see the 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#

Emilia Aguilar · Answer

undefined To find the sum of the series, use the formula for the sum of an arithmetic series: $ S = \frac{n}{2}(a_1 + a_n) $, where $ S $ is the sum, $ n $ is the number of terms, $ a_1 $ is the first term, and $ a_n $ is the last term.

First, find the number of terms:
$ n = \frac{a_n - a_1}{D} + 1 $

For this series, $ a_1 = 4 $, $ a_n = 170 $, and $ D $ is changing from $ +1 $ to $ +2 $.

Calculate $ n $ using the formula:
$ n = \frac{170 - 4}{1} + 1 = 167 + 1 = 168 $

Now, use the formula for the sum of an arithmetic series:
$ S = \frac{168}{2}(4 + 170) = 84 \times 174 = 14616 $

So, the sum of the series is $ 14616 $.