C1 + 4C2 + 7C3 + ...... + (3N - 2) Cn = _______

Given:

\( C_{1} + 4C_{2} + 7C_{3} + …… (3n – 2)C_{n} \) \( \Rightarrow \sum_{n=1}^{n} (3r – 2)C_{r} \) \( \Rightarrow 3 * \sum_{n=1}^{n} rC_{r} – 2\sum_{n=1}^{n} C_{r} \) \( \Rightarrow 3 * n \sum_{n=1}^{n} n – 1_{C_{r-1}} – 2\sum_{n=1}^{n} C_{r} \) \( \Rightarrow 3n\sum_{n=1}^{n} C_{k} – 2\sum_{n=1}^{n} C_{r} \) \( \Rightarrow 3n * (2^{n}-1) – 2 (2^{n}-1) \) \( \Rightarrow 2^{n}-1(3^{n}- 4) + 2 \)

Therefore, \( C_{1} + 4C_{2} + 7C_{3} + …… (3n – 2)C_{n} = 2^{n}-1(3^{n}- 4) + 2 \)

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