Tổng 1 + 11 + 101 + 1001 + ...... + 100...01 (12 số hạng) bằng:

A. \(\frac{{{{10}^{11}} + 107}}{9}\).

B. \(\frac{{{{10}^{12}} + 98}}{9}\).

C. \(\frac{{{{10}^{12}} + 107}}{9}\).

D. \(\frac{{{{10}^{11}} + 98}}{9}\).

Đáp án đúng là: B

Ta có 1 + 11 + 101 + 1001 + ...... + 100...01

= 1 + (10 + 1) + (100 + 1) + (1000 + 1) + ... + (100...0 + 1)

= 12 + (10 + 100 + 1000 + ... + 100...0)

= 12 + (10 + 10² + 10³ + ... + 10¹¹)

\( = 12 + \frac{{10\left( {1 - {{10}^{11}}} \right)}}{{1 - 10}}\)

\( = \frac{{108}}{9} + \frac{{{{10}^{12}} - 10}}{9}\)

\( = \frac{{{{10}^{12}} + 98}}{9}\).