Given-U-n-2n-1-n-even-3n-3-n-odd-The-value-of-U-3-U-6-U-7-U-10-U-11-U-14-U-27-U-28-

Question Number 128907 by bramlexs22 last updated on 11/Jan/21

Given U_{n} = {\begin{cases} 2 n + 1; n even \\ 3 n + 3; n odd \end{cases}

The value of U_{3} + U_{6} + U_{7} + U_{10} + U_{11} +

U_{14} + \dots + U_{27} + U_{28} = ?

Answered by Olaf last updated on 11/Jan/21

S = \underset{k = 3}{\sum^{k = 28}} U_{k}

S = \underset{n = 1}{\sum^{n = 13}} [U_{2 n + 1} + U_{2 n + 2}]

S = \underset{n = 1}{\sum^{n = 13}} [(3 (2 n + 1) + 3) + (2 (2 n + 2) + 1)]

S = \underset{n = 1}{\sum^{n = 13}} (10 n + 11)

S = 10 \frac{13 \times 14}{2} + 11 \times 13 = 910 + 143 = 1053

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