prove by mathmatical indiction 5+7+9+.....+(4n+1)=2n^2 +3n


Question and Answers Forum

All Questions Topic List
None Questions
Previous in All Question Next in All Question
Previous in None Next in None
Question Number 146102 by tabata last updated on 10/Jul/21

$p r o v e b y m a t h m a t i c a l i n d i c t i o n$ $5 + 7 + 9 + . . . . . + (4 n + 1) = 2 n^{2} + 3 n$
	Answered by gsk2684 last updated on 11/Jul/21

	$l e t S (n) : 5 + 7 + 9 + . . . + (4 n + 1) = 2 n^{2} + 3 n$ $b a s i c s t e p : i f n = 1 t h e n 5 = 2 {(1)}^{2} + 3 (1) i s t r u e$ $∴ S (1) i s t r u e$ $a s s u m p t i o n s t e p :$ $a s s u m e S (k) i s t r u e$ $5 + 7 + 9 + . . . + (4 k + 1) = 2 k^{2} + 3 k$ $i n d u c t i v e s t e p :$ $t o p r o v e S (k + 1) i s t r u e$ $5 + 7 + 9 + . . . + (4 k + 1) + (4 (k + 1) + 1)$ $= 2 k^{2} + 3 k + (4 k + 4 + 1)$ $= 2 k^{2} + 7 k + 5$ $= 2 (k^{2} + 2 k + 1) + 3 (k + 1)$ $= 2 {(k + 1)}^{2} + 3 (k + 1)$ $∴ S (k + 1) i s t r u e$ $∴ B y p r i n c i p l e o f f i n i t e m a t h e m a t i c a l$ $i n d u c t i o n g i v e n s t s t e m e n t$ $S (n) i s t r u e \forall n \in N$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum