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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} + \dots + {(2 n)}^{3} = 2 n^{2} {(n + 1)}^{2}

Answered by bemath last updated on 30/Jul/20

(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} + \dots + {(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} + \dots + {(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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