Question #136808
Compute the values of the sums below.
(a)5∑(k=1) (k+ 1)
(b)4∑(j=0) (-2) ^j
(c) 10∑(i=1) 3
(d) 8∑(j=0) ( 2^(j+1) -2^j )
1
Expert's answer
2020-10-15T17:36:27-0400

(a) 5k=1+(k+1)=5limn+k=1n(k+1)=5limn+(n+2)(n+1)2=+5\sum_{k=1}^{+\infty}(k+1)=5\, lim_{n\rightarrow+\infty}\sum_{k=1}^{n}(k+1)=5\, lim_{n\rightarrow+\infty}\frac{(n+2)(n+1)}{2}=+\infty

(b) 4j=0+(2)j=4limn+j=0n(2)j=limn+(2)n+113=4\sum_{j=0}^{+\infty}(-2)^j=4\, lim_{n\rightarrow+\infty}\sum_{j=0}^{n}(-2)^j\, =lim_{n\rightarrow+\infty}\frac{(-2)^{n+1}-1}{-3}=\infty

(c) 10i=1+3=+10\sum_{i=1}^{+\infty}3=+\infty

(d) 8j=0+(2j+12j)=8limn+j=0n(2j+12j)=8\sum_{j=0}^{+\infty}(2^{j+1}-2^{j})=8\,\,lim_{n\rightarrow+\infty}\sum_{j=0}^{n}(2^{j+1}-2^{j})=

=8limn+(2n+11)=+=8\,\,lim_{n\rightarrow+\infty}(2^{n+1}-1)=+\infty


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Comments

Assignment Expert
16.10.20, 01:12

Dear Promise Omiponle, final answers to the new question are (a) 20, (b) 11, (c) 30, (d) 511. If you need more details in a solution of the new question, please submit this new question with a help of the panel taking the previous recommendations into account.

Assignment Expert
16.10.20, 01:09

Dear Promise Omiponle, questions were incorrectly typed and coud be interpreted in different ways. You can write an additional description in questions if any confusion may occur there. In a new problem if you skip the limit sign, sums of the finite number of terms can be used from the previous solutions, but the final answer to a new question will differ surely.

Promise Omiponle
16.10.20, 00:45

Uh you guys actually made a mistake here. The numbers to the left of the sum from what I typed were meant to be the numbers on top. I put them on the left since well I had no way of putting them on top. I’m guessing this obviously can’t be done soon enough. The assignment is due Friday night at midnight. Well no worries I solved it myself because it’s no big deal. Wish y’all could get all the other discrete math problems I sent before tomorrow though.

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