Question

Sum of series
3+ 33+ 333+……to n terms

Answer 1

3+33+333+3333+33333+333333+3333333+333333333+3333333333

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Answer 2

Answer:

$\frac{10(10^n - 1)}{27 }$ - $\frac{n}{3}$

Step-by-step explanation:

3+ 33+ 333+……to n terms

= ($\frac{3}{3}$) * (3+ 33+ 333+……to n terms)

= ($\frac{1}{3}$) * (9+ 99+ 999+……to n terms)

= ($\frac{1}{3}$) * [ ( 10-1) + (100-1) + (1000-1) + ..... n terms ]

= ($\frac{1}{3}$) * [ ( 10¹-1) + (10²-1) + (10³-1) + ..... (10ⁿ-1) ]

=  ($\frac{1}{3}$) * [ ( 10¹ +10² + 10³ + ..... +10ⁿ) - (1+1+1+... n terms) ]

=  ($\frac{1}{3}$) * [ ( 10¹ +10² + 10³ + ..... +10ⁿ) - n ]

=  ($\frac{1}{3}$) * [ ( $\frac{10\times(10^n - 1)}{10-1 }$) - n ]

= ($\frac{1}{3}$) * [ ( $\frac{10\times(10^n - 1)}{9 }$) - n ]

=  $\frac{10(10^n - 1)}{27 }$ - $\frac{n}{3}$