Question

what are the next 2 term1, 11, 111, 1111 , _, _,

Answer 1

Consider that the given series takes the general form,

$a_n=1^{}+10+10^2+10^3+\cdots+10^n$

Observe that the expression is in geometric progression with first term 1 and common ratio 10, so applying the formula for the sum of geometric progression, the expression becomes,

$\begin{gathered} a_n=\frac{a(r^n-1)}{r-1} \\ a_n=\frac{1(10^n-1)}{10-1} \\ a_n=\frac{1}{9}\mleft(10^n-1\mright) \end{gathered}$

Then the 5th term is given by,

$\begin{gathered} a_5=\frac{1}{9}\mleft(10^5-1\mright) \\ a_5=\frac{1}{9}(10000^{}0-1) \\ a_5=\frac{1}{9}(99999) \\ a_5=11111 \end{gathered}$

Solve for the 6th term as,

$\begin{gathered} a_6=\frac{1}{9}\mleft(10^6-1\mright) \\ a_6=\frac{1}{9}(10000^{}00-1) \\ a_6=\frac{1}{9}(999999) \\ a_6=111111 \end{gathered}$

Thus, the next two terms are 11111, and 111111 respectively.