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Question 1
11866

To prove by mathematical induction that \(a + (a + d) + (a + 2d) + .... + (a + (n - 1)d) = \frac{n}{2}(2a + (n - 1)d)\) it is assumed that \(a + (a + d) + (a + 2d) + .... + (a + (k - 1)d) = \frac{k}{2}(2a + (k - 1)d)\). Then it must be proved that \(\dfrac{k}{2}(2a + (k - 1)d) + (a + kd) = \)

A.

\(\dfrac{{(2k + 1)(a + kd)}}{2}\)

B.

\(\dfrac{{(k + 1)(2a + (k + 1)d)}}{2}\)

C.

\(\dfrac{{(k + 1)(a + kd)}}{2}\)

D.

\(\dfrac{{(k + 1)(2a + kd)}}{2}\)

\(\dfrac{{(k + 1)(2a + kd)}}{2}\)

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