$2. C_{0} + 5. C_{1} + 8. C_{2} + . . . + (3 n + 2) C_{n} = (3 n + 4) 2^{n - 1}$

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Solution

$2 C_{0} + 5 C_{1} + 8 C_{2} + . . . . . . . . . . . . . (3 n + 4) C_{n}$
= $\sum_{r = 0}^{n} (3 r + 2) C_{r}$
= $\sum_{r = 0}^{n} 3 r . C_{r} + 2 C_{r}$ as for r=0 $3 r C_{r}$ would be 0
= $\sum_{r = 1}^{n} 3 r . \frac{n}{r} .^{n - 1} C_{r - 1} + \sum_{r = 0}^{n} 2. C_{r}$
= $\sum_{r = 1}^{n} 3 n .^{n - 1} C_{r - 1}$ + $\sum_{r = 0}^{n} 2. C_{r}$ As = $2^{n}$ = $^{n} C_{0} +^{n} C_{1} +^{n} C_{2} . . . . . . . . . . . +^{n} C_{n}$
.= $3 n {.2}^{n - 1} + {2.2}^{n}$
= $3 n {.2}^{n - 1} + {4.2}^{n - 1}$
= $(3 n + 4) {.2}^{n - 1}$

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