The sum of 2 + 22 + 222 + 2222 + ..... upto n terms is

When each term of a series is a repeated-digit number such as d, dd, ddd, ... (so the k-th term equals d × (10k − 1)/9), the standard method is to factor out d/9, rewrite every term as a power of 10 minus 1, split the result into a geometric part and a constant part, and apply the geometric-progression sum formula a(rn − 1)/(r − 1) to the power-of-10 part.

1. Let S = 2 + 22 + 222 + ... up to n terms.

2. Factor 2 out of every term: S = 2(1 + 11 + 111 + ... up to n terms).

3. Multiply and divide by 9 so each bracket term becomes a power of 10 minus 1: S = (2/9)(9 + 99 + 999 + ... up to n terms) = (2/9)[(10 − 1) + (102 − 1) + ... + (10n − 1)].

4. Separate the sum into a power-of-10 part and a constant part: S = (2/9)[(10 + 102 + ... + 10n) − (1 + 1 + ... + 1)], with n ones in the second bracket.

5. The power-of-10 part is a geometric progression with first term 10, common ratio 10, and n terms, so its sum is 10(10n − 1)/(10 − 1) = (10/9)(10n − 1).

6. The second part is n ones added together, which equals n.

7. Substitute both results back: S = (2/9){(10/9)(10n − 1) − n}.

Independent check with n = 2: the direct sum is 2 + 22 = 24. Substituting n = 2 into (2/9){(10/9)(102 − 1) − 2} gives (2/9){(10/9)(99) − 2} = (2/9)(110 − 2) = (2/9)(108) = 24, which agrees with the direct addition.

Therefore, S = (2/9){(10/9)(10n − 1) − n} for the sum of n terms of this series.