Question

The value of $2^2 + 22^2 + 222^2 ...$ upto n terms = $\frac{4}{81}[\frac{100(10^{2n}-1)}{99}-(n+2a)]$

Answer 1

False

Answer 2

The $k$-th term of the series is the number formed by repeating the digit 2, $k$ times. This can be written as:

$a_k = 2 \times \underbrace{11\dots1}_{k \text{ times}}$

The number $\underbrace{11\dots1}_{k \text{ times}}$ is a repunit, which can be expressed as $\frac{10^k - 1}{9}$.

So, the $k$-th term is $a_k = 2 \times \frac{10^k - 1}{9} = \frac{2}{9}(10^k - 1)$.

We need to find the sum of the squares of these terms up to $n$ terms:

$S_n = \sum_{k=1}^n a_k^2 = \sum_{k=1}^n \left(\frac{2}{9}(10^k - 1)\right)^2$

$S_n = \sum_{k=1}^n \frac{4}{81}(10^k - 1)^2$

$S_n = \frac{4}{81} \sum_{k=1}^n (10^{2k} - 2 \cdot 10^k + 1)$

We can separate the sum into three parts:

$S_n = \frac{4}{81} \left( \sum_{k=1}^n 10^{2k} - 2 \sum_{k=1}^n 10^k + \sum_{k=1}^n 1 \right)$

Let's evaluate each sum:

1. $\sum_{k=1}^n 1 = n$

2. $\sum_{k=1}^n 10^k = 10^1 + 10^2 + \dots + 10^n$. This is a geometric series with the first term $a = 10$, common ratio $r = 10$, and $n$ terms. The sum is $\frac{a(r^n - 1)}{r - 1} = \frac{10(10^n - 1)}{10 - 1} = \frac{10}{9}(10^n - 1)$.

3. $\sum_{k=1}^n 10^{2k} = \sum_{k=1}^n (10^2)^k = \sum_{k=1}^n 100^k = 100^1 + 100^2 + \dots + 100^n$. This is a geometric series with the first term $a = 100$, common ratio $r = 100$, and $n$ terms. The sum is $\frac{a(r^n - 1)}{r - 1} = \frac{100(100^n - 1)}{100 - 1} = \frac{100}{99}(10^{2n} - 1)$.

Substitute these sums back into the expression for $S_n$:

$S_n = \frac{4}{81} \left( \frac{100}{99}(10^{2n} - 1) - 2 \cdot \frac{10}{9}(10^n - 1) + n \right)$

$S_n = \frac{4}{81} \left( \frac{100(10^{2n} - 1)}{99} - \frac{20(10^n - 1)}{9} + n \right)$

Now, let's compare this derived formula with the given formula:

Given formula: $S_n = \frac{4}{81}[\frac{100(10^{2n}-1)}{99}-(n+2a)]$

Let's try to match the terms inside the square brackets. The first term $\frac{100(10^{2n}-1)}{99}$ matches.

The remaining terms in our derived formula are $-\frac{20(10^n - 1)}{9} + n$.

The remaining term in the given formula is $-(n+2a)$.

For the formulas to be equal, we would need:

$-\frac{20(10^n - 1)}{9} + n = -(n+2a)$

Since the value of 'a' depends on 'n', 'a' is not a constant. The structure of the derived formula involves a term with $10^n$, while the given formula has a term linear in $n$ (plus a constant). This indicates that the given formula is incorrect.

Therefore, the given formula is incorrect.