prove by mathematical induction 2^3 +4^3 +6^3 +8^3 +...+(2n)^3 = 2n^2 (n+1)^2


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Question Number 105605 by bobhans last updated on 30/Jul/20

$p r o v e b y m a t h e m a t i c a l i n d u c t i o n$ $2^{3} + 4^{3} + 6^{3} + 8^{3} + . . . + {(2 n)}^{3} = 2 n^{2} {(n + 1)}^{2}$
	Answered by bemath last updated on 30/Jul/20
	$(1)P(1) = { ((lhs = 8)),((rhs=2.(2)^2 =8)) :} (true) (2) let P(k) is true then 2^3 +4^3 +6^3 +...+(2k)^3 =2k^2 (k+1)^2 (3) put n = k+1 lhs : 2^3 +4^3 +6^3 +...+(2k)^3 +(2(k+1))^3 = 2k^2 (k+1)^2 +2^3 (k+1)^3 = 2(k+1)^2 {k^2 +4(k+1)}= 2(k+1)^2 {k^2 +4k+4}= 2(k+1)^2 (k+2)^2 rhs : 2(k+1)^2 (k+1+1)^2 = 2(k+1)^2 (k+2)^2 true ⇒ QED$
	$(1) P (1) = {\begin{cases} l h s = 8 \\ r h s = 2 . {(2)}^{2} = 8 \end{cases} (t r u e)$ $(2) l e t P (k) i s t r u e t h e n$ $2^{3} + 4^{3} + 6^{3} + . . . + {(2 k)}^{3} = 2 k^{2} {(k + 1)}^{2}$ $(3) p u t n = k + 1$ $l h s : 2^{3} + 4^{3} + 6^{3} + . . . + {(2 k)}^{3} + {(2 (k + 1))}^{3} =$ $2 k^{2} {(k + 1)}^{2} + 2^{3} {(k + 1)}^{3} =$ $2 {(k + 1)}^{2} {k^{2} + 4 (k + 1)} =$ $2 {(k + 1)}^{2} {k^{2} + 4 k + 4} =$ $2 {(k + 1)}^{2} {(k + 2)}^{2}$ $r h s : 2 {(k + 1)}^{2} {(k + 1 + 1)}^{2} =$ $2 {(k + 1)}^{2} {(k + 2)}^{2}$ $t r u e \Rightarrow Q E D$
	Answered by Dwaipayan Shikari last updated on 30/Jul/20

	$W i t h o u t M a t h e m a t i c a l I n d u c t i o n$ $T_{n} = {(2 n)}^{3}$ $Σ T_{n} = 8 Σ n^{3}$ $\underset{n = 1}{\sum^{n}} T_{n} = 8 {(\frac{n (n + 1)}{2})}^{2} = 2 n^{2} {(n + 1)}^{2}$
	Commented by JDamian last updated on 30/Jul/20

	$‘ ‘ F r o m n = 1 u p t o n^{″} ?$
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