If sumlimits_{i = 1}^n i = frac{n(n + 1)}{2}, | vedclass

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(A) We are given the sum $\sum\limits_{i = 1}^n (3i - 2)$.Using the linearity property of summation,we can write:$\sum\limits_{i = 1}^n (3i - 2) = 3\s

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$\frac{n(3n - 1)}{2}$

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(A) We are given the sum $\sum\limits_{i = 1}^n (3i - 2)$.Using the linearity property of summation,we can write:$\sum\limits_{i = 1}^n (3i - 2) = 3\sum\limits_{i = 1}^n i - \sum\limits_{i = 1}^n 2$Since $\sum\limits_{i = 1}^n i = \frac{n(n + 1)}{2}$ and $\sum\limits_{i = 1}^n 2 = 2n$,we have:$= 3 \left[ \frac{n(n + 1)}{2} \right] - 2n$$= \frac{3n^2 + 3n - 4n}{2}$$= \frac{3n^2 - n}{2}$$= \frac{n(3n - 1)}{2}$.

If $\sum\limits_{i = 1}^n i = \frac{n(n + 1)}{2}$,then $\sum\limits_{i = 1}^n (3i - 2) = $

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