An algebra problem by anshu garg

$\begin{aligned} S & = 7+77+777+...+7777777777 \\ & = 7({\color{#3D99F6}\underbrace{1}_1} + 1{\color{#3D99F6}\underbrace{1}_2}+11 {\color{#3D99F6}\underbrace{1}_3}+...+111111111 {\color{#3D99F6}\underbrace{1}_{10}}) \\ & = \sum_{n=0}^9 7(10-n)10^{n+1} \\ & = 7 \sum_{n=0}^9 \left(10^{n+2}-n10^{n+1} \right) \\ & = 700 \sum_{n=0}^9 10^n - 70 \color{#3D99F6} \sum_{n=1}^9 n10^n & \small \color{#3D99F6} \text{See note.} \\ & = 700 \cdot \frac {10^{10}-1}{10-1} - 70\cdot \color{#3D99F6} 9876543210 \\ & = 700 \cdot 1,111,111,111 - 691,358,024,700 \\ & = 777,777,777,700 - 691,358,024,700 \\ & = \boxed{86,419,753,000} \end{aligned}$

Note:

$\begin{aligned} S_1 & = \sum_{n=1}^9 n10^n \\ & = 10+2(10^2) + 3(10^3)+...+9(10^9) \\ 10S_1 & = 10^2+2(10^3) + 3(10^4)+...+9(10^{10}) \\ (1-10)S_1 & = 10 + 10^2+10^3 + ... + 10^9 - 9(10^{10}) \\ -9S_1 & = 10 \cdot \frac {10^9-1}{10-1} - 9(10^{10}) \\ & = \frac {10^{10}-10-81(10^{10})}9 \\ \implies S_1 & = \frac {8\times 10^{11}-10}{81} = 9876543210 \end{aligned}$

Divide each term by 7, the terms are then: 1, 11, 111.... 1,111,111,111). The first 9 terms will add up to 123456789 (since 9 of the terms have a 1's digit, 8 a ten's digit, etc.). The last term is merely 10 1's. Add those two numbers and you get 1234567900. Multiply by 7 (that we initially divided the terms by) to get 8641975300.

7(1+11+111.................................................1111111111)

$\frac{7}{9}$ (9)(1+11+111+1111..............................................10 terms)

$\frac{7}{9}$ (9+99+999+9999..............................10 terms)

$\frac{7}{9}$ ([10-1]+[100-1]+[1000-1]................................10 terms)

Now Applying G.P. Formula and simplifying

=86419753000