Find the indicated sum.
-$$\sum _ { n = 1 } ^ { 42 } ( - 3 n - 3 )$$

Question

Quizplus · Accepted Answer

The given sum can be simplified by distributing the summation over each term: $\sum _ { n = 1 } ^ { 42 } ( - 3 n - 3 ) = \sum _ { n = 1 } ^ { 42 } ( - 3 n ) + \sum _ { n = 1 } ^ { 42 } ( - 3 )$. The first part, $\sum _ { n = 1 } ^ { 42 } ( - 3 n )$, is a sum of an arithmetic series with the first term $a_1 = -3$, common difference $d = -3$, and $n = 42$ terms. The second part is simply $42$ times $-3$, which is $-126$. The sum of the first $n$ terms of an arithmetic series is given by $S_n = \frac{n}{2} [2a_1 + (n - 1)d]$. Substituting $n = 42$, $a_1 = -3$, and $d = -3$, we get $S_n = \frac{42}{2} [2(-3) + (42 - 1)(-3)] = 21[-6 - 123] = 21[-129] = -2709$. Adding the second part, $-126$, gives the total sum: $-2709 - 126 = -2835$.

Find the Indicated Sum ∑n=142(−3n−3)\sum _ { n = 1 } ^ { 42 } ( - 3 n - 3 )n=1∑42​(−3n−3)

Find the Indicated Sum $\sum _ { n = 1 } ^ { 42 } ( - 3 n - 3 )$