Question

prove that the sum of n terms of the sequence (9,99,999,.............) equals 10/9(10 to the power of n -1 )-n

Answer 1

well by the way -1 and -n are not powers

Answer 2

\[\underbrace{9+99+999+\cdots}_{n\quad times}=10-1+100-1+1000-1+\cdots=\]\[=(10+10^2+10+10^3+\cdots+10^n)-n=10\cdot\frac{10^n-1}{10-1}-n=\]\[=\frac{10}{9}(10^n-1)-n\]

Answer 3

snappy!

Answer 4

so he called the repeatative -1 as -n

Answer 5

because it said to n terms right