Question

sum the series 5+55+555+..... to n terms


answer show be short and error free

Answer 1

5(1+11+111...........n term)
5÷9(9+99+999...........n term)
5÷9[(10-1)+(

Answer 2

We have

$5 + 55 + 555 + .... \: to \: n \: terms \\ \\ taking \: common \: 5 \: from \: each \: term \\ \\ = 5 \times [1 + 11 + 111 + .... \: to \: n \: terms]\\ \\ = \frac{5}{9} \times [9 + 99 + 999 + .... \: to \: n \: terms]\\ \\ = \frac{5}{9} \times [(10 - 1) + ({10}^{2} - 1) + ({10}^{3} - 1] ) + ...to \: n \: terms]\\ \\ = \frac{5}{9} \times[ (10 + {10}^{2} + {10}^{3} + ...n \: terms) - n \times 1] \\ \\ = \frac{5}{9} \times [ \frac{10 \times ( {10}^{n} - 1) }{(10 - 1)} - n] \\ \\ = \frac{5}{9} \times [ \frac{ {10}^{n + 1} - 10 - 9n}{9} ] \\ \\ = \frac{5}{81} \times ( {10}^{n + 1} - 9n - 10)$

⭐Hope it may help you⭐