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

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(A) Given the summation $\sum_{i = 1}^n (3i - 2)$.Using the linearity property of summation,we can write:$\sum_{i = 1}^n (3i - 2) = 3 \sum_{i = 1}^n i

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

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(A) Given the summation $\sum_{i = 1}^n (3i - 2)$.Using the linearity property of summation,we can write:$\sum_{i = 1}^n (3i - 2) = 3 \sum_{i = 1}^n i - \sum_{i = 1}^n 2$.Since $\sum_{i = 1}^n i = \frac{n(n + 1)}{2}$ and $\sum_{i = 1}^n 2 = 2n$,we substitute these values:$= 3 \left[ \frac{n(n + 1)}{2} \right] - 2n$.$= \frac{3n^2 + 3n}{2} - 2n$.$= \frac{3n^2 + 3n - 4n}{2}$.$= \frac{3n^2 - n}{2} = \frac{n(3n - 1)}{2}$.

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

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