Question

Find the sum of first n terms of :
i) $4+44+444+...$
ii) $0.7+0.77+0.777+...$

Answer 1

i)

4+44+444+....... upto n terms.

4(1+11+111+........upto n terms)

$\frac{4}{9}$(9+99+999+.........upto n terms)

$\frac{4}{9}\left[\right(10-1)+({10}^2-1)+$...... upto n terms]

$\frac{4}{9}[10+{10}^2+.........+{10}^n$}-(1+1+1+....n times)]

$\frac{4}{9}10\left[\frac{({10}^n-1)}{(10-1)}\right]-n\left(1\right)$

$\frac{4}{9}\left[\frac{10}{9}\right({10}^n-1)-n]$...................

ii)

Sum = 0.7 + 0.77 + 0.777 + …. up to n terms

= $\frac{7}{9}$(0.9 + 0.99 + 0.999 + … up to n terms)

= $\frac{7}{9}$[(1 – 0.1) + (1 – 0.01) + (1-0.001) + … up to n terms]

= $\frac{7}{9}\left[\right(1+1+1\dots uptonterms)–(\frac{1}{10}+\frac{1}{100}+\frac{1}{1000}+\dots uptonterms\left)\right]$

= $\frac{7}{9}[n–0.1\times \frac{(1–(0.1\left)n\right)}{(1–0.1)}]$

= $\frac{7}{81}[9n–1+{10}^{-n}]$