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 + ……. + 10n – 1)²
1, 10², 10³, ……. + 10n – 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}