- #1

courtrigrad

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(1) Let's say you have the progression [itex] 5 + 12 + 21 + ... + 1048675 [/itex] and you want to find the sum of 20 terms. I know that the kth term is given by [itex] 2^{k} +3 + 5(k-1) [/itex]. So would I treat the [itex] 2^{k} [/itex] terms separately from the [itex] 3 + 5(k-1) [/itex] terms? Would it be [itex] 2 + 4 + 8 + 16 +... + 2^{n} [/itex] and [itex] 3 + 8 + 13 + ... + (3+5(k-1)) [/itex]. Would the total sum be [itex] \frac{2 - 2(2)^{n}}{-1} + \frac{n}{2}(3+ (3+5(k-1)) [/itex]?

[tex] 2097248 + 1010 [/tex]

(2) [tex] 3 + 10 + 25 + ... + 39394 [/tex] and you want to find sum of first 10 terms. I know that the kth term is [itex] 2 \times 3^{k-1} + 1 + 3(k-1) [/itex] Would i do the same thing and treat the [itex] 3^{k-1} [/itex] and [itex] 1 + 3(k-1) [/itex] separately?

Thanks

[tex] 2097248 + 1010 [/tex]

(2) [tex] 3 + 10 + 25 + ... + 39394 [/tex] and you want to find sum of first 10 terms. I know that the kth term is [itex] 2 \times 3^{k-1} + 1 + 3(k-1) [/itex] Would i do the same thing and treat the [itex] 3^{k-1} [/itex] and [itex] 1 + 3(k-1) [/itex] separately?

Thanks

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