Rewrite the series as an equivalent series with the new index of summation.
-$$\sum _ { i = 1 } ^ { 7 } ( 9 i ) = \sum _ { k = 3 } ^ { ? } ( ? )$$
A) $\sum _ { k = 3 } ^ { 9 } [ 9 ( k - 2 ) ]$
B) $\sum _ { k = 3 } ^ { 9 } [ 9 ( k + 2 ) ]$
C) $\sum _ { k = 3 } ^ { 9 } [ 9 ( k - 3 ) ]$
D) $\sum _ { k = 3 } ^ { 9 } [ 9 ( k + 3 ) ]$

Question

Quizplus · Accepted Answer

To change the index of summation from $i = 1$ to $k = 3$, we substitute $i = k - 2$. This changes the expression inside the sum to $9(k - 2)$. The upper limit changes from $i = 7$ to $k = 9$ to maintain the same number of terms in the series.

Rewrite the Series as an Equivalent Series with the New ∑i=17(9i)=∑k=3?(?)\sum _ { i = 1 } ^ { 7 } ( 9 i ) = \sum _ { k = 3 } ^ { ? } ( ? )i=1∑7​(9i)=k=3∑?​(?)

Rewrite the Series as an Equivalent Series with the New $\sum _ { i = 1 } ^ { 7 } ( 9 i ) = \sum _ { k = 3 } ^ { ? } ( ? )$