Question

Find the sum 9+99+999+9999....+99.....9.

Answer 1

I assume you want answer up to n terms i.e.

9+99+999+9999+99999+……. n terms

Let S= 9+99+999+9999+99999+……. n terms

S= (10–1)+(100–1)+(1000–1)+(1000–1)…..n terms

S= {(10+100+1000+…..n terms)-(1+1+1+1+…..n terms)}

S= {(10+100+1000+…..n terms)-n}

Therefore first bracket becomes a GP.

S= 10{ 10^(n-1) - 1}/(10–1) - n

S= [10{ 10^(n-1) - 1}/9] - n

Answer 2

infinite.............