Find-the-sum-of-n-terms-of-the-series-S-n-1-22-333-4444-

Question Number 121235 by ZiYangLee last updated on 06/Nov/20

Find the sum of n terms of the series

S_{n} = 1 + 22 + 333 + 4444 + \dots \dots

Answered by Ar Brandon last updated on 06/Nov/20

S_{n} = 1 + 22 + 333 + 4444 + \cdot \cdot \cdot

= 1 (10^{0}) + 2 (10^{1} + 10^{0}) + 3 (10^{2} + 10^{1} + 10^{0})

+ 4 (10^{3} + 10^{2} + 10^{1} + 10^{0})

= 10^{0} (1 + 2 + 3 + \cdot \cdot \cdot + n) + 10^{1} (2 + 3 + 4 + \cdot \cdot \cdot + n)

+ 10^{2} (3 + 4 + 5 + \cdot \cdot \cdot + n) + \cdot \cdot \cdot + 10^{n - 1} (n)

= \frac{n (n + 1)}{2} + \frac{10 (n - 1) (2 + n)}{2} + \frac{10^{2} (n - 2) (3 + n)}{2} + \cdot \cdot \cdot + \frac{10^{n - 1} (n)}{1}

T o b e c o n t i n u e d \dots

Answered by Dwaipayan Shikari last updated on 06/Nov/20

F i r s t t e r m = \frac{10^{n} n}{9} - \frac{n}{9} n = 1

s e c o n d t e r m = \frac{10^{2} . 2}{9} - \frac{2}{9}

t h i r d t e r m = \frac{10^{3} . 3}{9} - \frac{3}{9}

T_{n} = \frac{10^{n} . n}{9} - \frac{n}{9}

\underset{n = 1}{\sum^{n}} T_{n} = \underset{n = 1}{\sum^{n}} \frac{10^{n} . n}{9} - \underset{n = 1}{\sum^{n}} \frac{n}{9}

S = \sum^{n} n . 10^{n}

S = 10 + 2 . 10^{2} + \dots + n 10^{n}

- 10 S = - 10^{2} - \dots . - (n - 1) 10^{n} - n 10^{n + 1}

- 9 S = 10 + 10^{2} + \dots + 10^{n} - n 10^{n + 1}

S = \frac{n}{9} 10^{n + 1} - \frac{1}{81} (10^{n + 1} - 10)

\sum^{n} T_{n} = \frac{1}{9} (\frac{n}{9} 10^{n + 1} - \frac{1}{81} 10^{n + 1} + \frac{10}{81}) - \sum^{n} \frac{n}{9}

= \frac{1}{9} (\frac{n}{9} 10^{n + 1} - \frac{1}{81} 10^{n + 1} + \frac{10}{81}) - \frac{n (n + 1)}{18}

Commented by Dwaipayan Shikari last updated on 06/Nov/20

C h e c k i n g

n = 1

\frac{1}{9} (\frac{100}{9} - \frac{100}{81} + \frac{10}{81}) - \frac{2}{18} = \frac{10}{9} - \frac{1}{9} = 1 (t r u e)