determinant-determinant-2-424-44244-4442444-n-terms-

Question Number 175548 by Rasheed.Sindhi last updated on 02/Sep/22

\begin{matrix} \overset{\underline{\overset{―}{∣ ∙ ∣}}}{{\begin{matrix} \underset{}{\overset{}{2 + 424 + 44244 + 4442444 + \cdot \cdot \cdot n t e r m s = ?_{}^{}}} \end{matrix}}_{}^{}} \end{matrix}

Commented by infinityaction last updated on 02/Sep/22

(2 + 2) + (424 + 20) + (44244 + 200) + \dots n t e r m

- (2 + 20 + 200 + \dots + n t e r m)

S = {4 + 444 + 44444 + \dots n t e r m

- 2 (1 + 10 + 100 + . . n t e r m)}

\underset{Ψ}{\underset{⏟}{4 + 444 + 44444 + \dots . . n t e r m}} - \frac{2}{9} (10^{n} - 1)

Ψ = 4 (1 + 111 + 11111 + \dots n t e r m)

9 Ψ = 4 (9 + 999 + \dots n t e r m)

Ψ = \frac{4}{9} [{10 - 1} + {10^{3} - 1} + \dots n t e r m]

Ψ = \frac{4}{9} [\frac{10}{99} \times (10^{2 n} - 1) - n]

S = Ψ - \frac{2}{9} (10^{n} - 1)

S = \frac{4}{9} [\frac{10}{99} \times (10^{2 n} - 1) - n] - \frac{2}{9} (10^{n} - 1)

Commented by Rasheed.Sindhi last updated on 03/Sep/22

∙ \cap i \subset\in!

∙ T h a n k y o u s i r!

Commented by infinityaction last updated on 03/Sep/22

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Commented by peter frank last updated on 05/Sep/22

thanks

Answered by Rasheed.Sindhi last updated on 04/Sep/22

AnOther way \dots

2 + 424 + 44244 + 4442444 + \cdot \cdot \cdot n t e r m s

= (\underset{n - 1 t e r m s}{\underset{⏟}{404 + 44044 + 4440444 + \dots}}) + (\underset{n t e r m s}{\underset{⏟}{2 + 20 + 200 + \dots}})

= \frac{4}{9} (\underset{n - 1 t e r m s}{\underset{⏟}{909 + 99099 + 9990999 \dots}}) + \frac{2 (10^{n} - 1)}{(10 - 1)}

= \frac{4}{9} (\underset{n - 1 t e r m s}{\underset{⏟}{1000 - 91 + 100000 - 901 + 10000000 - 9001 \dots}}) + \frac{2 (10^{n} - 1)}{9}

= \frac{4}{9} (\underset{n - 1 t e r m s}{\underset{⏟}{10^{3} - 91 + 10^{5} - 901 + 10^{7} - 9001 \dots}}) + \frac{2 (10^{n} - 1)}{(10 - 1)}

= \frac{4}{9} {(\underset{n - 1 t e r m s}{\underset{⏟}{10^{3} + 10^{5} + 10^{7} + \dots}}) - (\underset{n - 1 t e r m s}{\underset{⏟}{90 + 900 + 9000 + \dots}}) - (n - 1)} + \frac{2 (10^{n} - 1)}{(10 - 1)}

= \frac{4}{9} {\frac{10^{3} (100^{n - 1} - 1)}{100 - 1} - \frac{90 (10^{n - 1} - 1)}{10 - 1} - n + 1} + \frac{2 (10^{n} - 1)}{(10 - 1)}

= \frac{4}{9} {\frac{10^{3} (100^{n - 1} - 1)}{99} - \frac{90 (10^{n - 1} - 1)}{9} - n + 1} + \frac{2 (10^{n} - 1)}{9}

= \frac{4}{9} {\frac{10^{3} (100^{n - 1} - 1) - 990 (10^{n - 1} - 1) - 99 n + 99}{99}} + \frac{2 (10^{n} - 1)}{9}

= \frac{4}{9} {\frac{10^{2 n + 1} - 990 \cdot 10^{n - 1} - 99 n + 89}{99}} + \frac{2 (10^{n} - 1)}{9}

= \frac{4}{9} {\frac{10 \cdot 10^{2 n} - 10 - 99 \cdot 10^{n} - 99 n + 99}{99}} + \frac{2 (10^{n} - 1)}{9}

= \frac{4}{9} {\frac{10}{99} (10^{2 n} - 1) - n} - \frac{2}{9} ((10^{n} - 1))

Commented by Tawa11 last updated on 15/Sep/22

Great sir

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