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Question Number 9471 by tawakalitu last updated on 09/Dec/16

prove by mathematical induction

a + (a + d) + (a + 2 d) + \dots + [a + (n - 1) d] = \frac{1}{2} n [2 a + (n - 1) d]

Answered by mrW last updated on 10/Dec/16

for n = 1 :

a \overset{!}{=} \frac{1}{2} \times 1 \times [2 a + (1 - 1) d] = a

for n + 1 :

a + (a + d) + (a + 2 d) + \dots + [a + (n - 1) d] + [a + (n + 1 - 1) d]

= \frac{1}{2} n [2 a + (n - 1) d] + [a + (n + 1 - 1) d]

= \frac{1}{2} n [2 a + (n - 1) d] + [a + nd]

= \frac{n [2 a + (n - 1) d] + 2 (a + nd)}{2}

= \frac{2 na + n (n - 1) d + 2 (a + nd)}{2}

= \frac{2 (n + 1) a + (n + 1) nd}{2}

= \frac{1}{2} (n + 1) [2 a + (n + 1 - 1) d]

\Rightarrow proved!

Commented by tawakalitu last updated on 10/Dec/16

Thanks so much sir . God bless you .