Use mathematical induction to prove that the statement a+(a+d)+(a+2d)+...+(a+(n−1)d)=(n/2)[2a+(n−1)d] is true for all natural numbers Any help please


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Question Number 199987 by Yakubu last updated on 11/Nov/23

$U s e m a t h e m a t i c a l i n d u c t i o n$ $t o p r o v e t h a t t h e s t a t e m e n t$ $a + (a + d) + (a + 2 d) + . . . + (a + (n - 1) d) = \frac{n}{2} [2 a + (n - 1) d]$ $is true for all natural numbers$ $A n y h e l p p l e a s e$
	Answered by Rasheed.Sindhi last updated on 12/Nov/23

	$p (n) : a + (a + d) + (a + 2 d) + . . . + (a + (n - 1) d) = \frac{n}{2} [2 a + (n - 1) d]$ $∙ p (1) : a + (1 - 1) d = \frac{1}{2} (2 a + (1 - 1) d)$ $a = a$ $p (1) i s t r u e$ $∙ l e t p (k) i s t r u e :$ $p (k) : a + (a + d) + (a + 2 d) + . . . + (a + (k - 1) d) = \frac{k}{2} [2 a + (k - 1) d]$ $. . . . + (a + (k - 1) d) + a + (\overset{―}{k + 1} - 1) d = \frac{k}{2} [2 a + (k - 1) d] + a + (k + 1 - 1) d$ $= k a + k (\frac{k - 1}{2}) d + a + k d$ $= a (k + 1) + (\frac{k^{2} - k}{2} + k) d$ $= a (k + 1) + (\frac{k^{2} - k + 2 k}{2}) d$ $= 2 a (\frac{k + 1}{2}) + (\frac{k^{2} + k}{2}) d$ $= 2 a (\frac{k + 1}{2}) + k (\frac{k + 1}{2}) d$ $= 2 a (\frac{k + 1}{2}) [2 a + k d]$ $= 2 a (\frac{k + 1}{2}) [2 a + (\overset{―}{k + 1} - 1) d]$ $∴ p (k + 1) i s t r u e w h e n p (k) i s t r u e$ $H e n c e p (n) i s t r u e \forall n \in N$
	Commented by Yakubu last updated on 12/Nov/23

	$t h a n k y o u s i r$
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