$C_{1} + 4 C_{2} + 7 C_{3} + . . . . . + (3 n - 2) C_{n} =$

{(3 n - 4) 2}^{n + 1}

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{(3 n - 4) 2}^{n - 1} + 2

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{(3 n - 4) 2}^{n}

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{(3 n - 4) 2}^{n - 1} + 1

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Solution

The correct option is A ${(3 n - 4) 2}^{n - 1} + 2$
$C_{1} + 4_{C_{2}} + 7_{C_{3}} + . . . . . . (3 n - 2) C_{n}$

$= \sum_{n = 1}^{n} (3 r - 2) C_{r}$
$= 3. \sum_{r = 1}^{n} r C_{r} - 2 \sum_{r = 1}^{n} C_{r}$

$= 3. n \sum_{r = 1}^{n} {n - 1}_{C_{r - 1}} - 2 \sum_{r = 1}^{n} C_{r}$
$= 3 n \sum_{r = 0}^{n - 1} C_{k} - 2 \sum_{r = 1}^{n} C_{r}$

$= 3 n . (2^{n - 1}) - 2 (2^{n} - 1)$

$= 2^{n - 1} (3 n - 4) + 2$ [B]

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