# Find the Sum of the Square of two Numbers With n Digits

## Sequence and series math problem

Find the sum of the squares of two numbers (666…6)² and (444…4)², Both numbers has n digits

## Solution

(666…6)² + (444…4)² = (6 × 111…1)² + (4 × 111…1)²

⇒ (666…6)² + (444…4)² = 52 × (111…1)²

⇒ (666…6)² + (444…4)² = 52 × (1 + 10 + 100 + 1000 + ……*n* terms)²

⇒ (666…6)² + (444…4)² = 52 × (1 + 10² + 10³ + …….* *+ 10^{n – 1})²

## 1, 10², 10³, …….* *+ 10^{n – 1} is a geometric series, we need to solve the series to find the sum of the square of two numbers

\begin{aligned} 1+10^2+10^3+....+10^{n-1}&=\frac{1 \times (10^{n}-1)}{10-1} \\ \\ \Rightarrow 1+10^2+10^3+....+10^{n-1}&= \frac{10^{n}-1}{9}\\ \\ \Rightarrow (1+10^2+10^3+....+10^{n-1})^2&= \frac{(10^{n}-1)^2}{81}\\ \end{aligned}

So (666…6)² + (444…4)² is

(666...6)^2+(444...4)^2=52 \times \frac{(10^{n}-1)^2}{81}

### Now, sum of the square is **(666…6)² + (444…4)² = (10***ⁿ* – 1)**²** × 52/81

*ⁿ*– 1)

**²**× 52/81