Find the partial sum. $\sum _ { n = 1 } ^ { 140 } ( - 4 n + 1 )$
A)-39,339
B)-38,781
C)-39,620
D)-39,903
E)-39,340

Question

Quizplus · Accepted Answer

The given sum can be simplified by distributing the summation over each term: $\sum _ { n = 1 } ^ { 140 } ( - 4 n + 1 ) = \sum _ { n = 1 } ^ { 140 } ( - 4 n ) + \sum _ { n = 1 } ^ { 140 } 1$. The first part is an arithmetic series with a common difference of -4, and the second part is simply adding 1 a total of 140 times. The sum of an arithmetic series can be found using the formula $S = \frac{n}{2} [2a + (n-1)d]$, where $n$ is the number of terms, $a$ is the first term, and $d$ is the common difference. For the first part, $a = -4$, $d = -4$, and $n = 140$, so the sum is $S_1 = \frac{140}{2} [2(-4) + (140-1)(-4)] = 70[-8 - 556] = 70[-564] = -39480$. The second part is simply $140 \times 1 = 140$. Adding these two results gives the total sum: $-39480 + 140 = -39340$.

Find the Partial Sum ∑n=1140(−4n+1)\sum _ { n = 1 } ^ { 140 } ( - 4 n + 1 )∑n=1140​(−4n+1)

Find the Partial Sum $\sum _ { n = 1 } ^ { 140 } ( - 4 n + 1 )$