Question

plz help me

Answer 1

\[\sum_{k=1}^{154}(-1)^{k+1}(\frac{ 2k+1 }{ k^{2}+k })\] express this as fraction

Answer 2

you mean as an explicit form?

Answer 3

yep

Answer 4

\[\sum_{k=1}^{154}(-1)^{k+1}(\frac{ 2k+1 }{ k^{2}+k })\]
\[\sum_{k=1}^{154}(-1)^{k+1}\frac{ 2k+1 }{ k(k+1) }\]
\[\sum_{k=1}^{154}(-1)^{k+1}\left(\frac{2}{k+1}+\frac{ 1 }{ k(k+1)}\right)\]
\[\sum_{k=1}^{154}(-1)^{k+1}\frac{2}{k+1}~+~\sum_{k=1}^{154}(-1)^{k+1}\frac{ 1 }{ k(k+1)}\]
\[2\sum_{k=1}^{154}(-1)^{k+1}\frac{1}{k+1}~+~\sum_{k=1}^{154}(-1)^{k+1}\frac{ 1 }{ k(k+1)}\]
this might help ...

Answer 5

we can do a partial fraction decomp on the last term to split it if need be

Answer 6

bro i dnt knw abt \[\sum_{}^{}\] and wht does the numbers beside this mean will u plz expln me frm the bgnning????@amistre64

Answer 7

let me work this out some more first
\[2\sum_{k=1}^{154}(-1)^{k+1}\frac{1}{k+1}~+~\sum_{k=1}^{154}(-1)^{k+1}\frac{ 1 }{ k(k+1)}\]
\[2\sum_{k=1}^{154}(-1)^{k+1}\frac{1}{k+1}~+~\sum_{k=1}^{154}(-1)^{k+1}\frac{ 1+k-k }{ k(k+1)}\]
\[2\sum_{k=1}^{154}(-1)^{k+1}\frac{1}{k+1}~+~
\sum_{k=1}^{154}(-1)^{k+1}\frac{ 1+k}{ k(k+1)}-\frac{ k }{ k(k+1)}\]
\[2\sum_{k=1}^{154}(-1)^{k+1}\frac{1}{k+1}~+~
\sum_{k=1}^{154}(-1)^{k+1}\frac{ 1}{ k}-\frac{ 1 }{ k+1}\]
\[2\sum_{k=1}^{154}(-1)^{k+1}\frac{1}{k+1}~+~
\sum_{k=1}^{154}(-1)^{k+1}\frac{ 1}{ k}-\sum_{k=1}^{154}(-1)^{k+1}\frac{ 1 }{ k+1}\]

Answer 8

a \(\sum\) is a greek letter Sigma (S) and is meant to convey the idea of a Sum of the parts from a starting value to an ending value\[\sum_{start}^{end}\]

Answer 9

from that last line, i see that we have "like" terms that we can work with ...

Answer 10

ok bt wt do the numbers beside the symbol mean??

Answer 11

the expression to the right of the symbol? that defines the rule that you will be operating on.

in this case there will be 154 iteration of the rule, and for each iteration you substitute in the "k" value for that iteration. Then you sum up all the iterations.

Answer 12

\[\left(2\sum_{k=1}^{154}(-1)^{k+1}\frac{1}{k+1}\right)~+~
\sum_{k=1}^{154}(-1)^{k+1}\frac{ 1}{ k}-\left(\sum_{k=1}^{154}(-1)^{k+1}\frac{ 1 }{ k+1}\right)\]

\[\sum_{k=1}^{154}(-1)^{k+1}\frac{1}{k+1}~+~
\sum_{k=1}^{154}(-1)^{k+1}\frac{ 1}{ k}\]
if we write out a few of the terms, we might be able to see a telescoping effect ....

Answer 13

can we substitute the value of k in the expression to the right of the symbol??? @amistre64

Answer 14

\[k=1~:~\cancel{\frac{1}{2}}+\frac{1}{1}\]
\[k=2~:~-\cancel{\frac{1}{3}}-\cancel{\frac{1}{2}}\]
\[k=3~:~\cancel{\frac{1}{4}}+\cancel{\frac{1}{3}}\]
\[k=4~:~-\cancel{\frac{1}{5}}-\cancel{\frac{1}{4}}\]...\[k=154~:~-\frac{1}{155}-\cancel{\frac{1}{154}}\]

total sum is just: 1 - 1/155 = 154/155

Answer 15

yes, that is what the start and end interval around the Sigma tells us to do

Answer 16

for k = 1,2,3,4,...,154: do (this rule), then add the results

Answer 17

@amistre64 olz expln frm whr 154 comes??

Answer 18

from the person who wrote the rule .... they wanted you to attempt the problem along the interval 1 to 154 .... its purely arbitrary values that the creator of the problem dreamed up

Answer 19

they tell you to start at k=1, and move along until k=154

Answer 20

ok wt did u do with the expressions which r to the right of the symbol plz expln????

Answer 21

i simplified it .... just algebra work

Answer 22

and what is brilliant.org?

Answer 23

ok did u multiply them???

Answer 24

no, i simplified them to a point where it became less obtrusive so that i could notice its telescoping nature

Answer 25

once a saw that, i could develop a plan to see how it works upon itself

Answer 26

i ddnt get u bro??

Answer 27

bro r u asking me the meaning of 'brilliant.org'???

Answer 28

someone suggested that you are posting content from another site .. i was just wondering.

Answer 29

i ddnt want to cheat bro bt i wntd to understand itz meaning

Answer 30

the premise is simple:

\[\sum_{k=0}^{k=N}f(k)=f(0)+f(1)+f(2)+f(3)+...+f(N)\]

Answer 31

the hard part is in trying to make f(k) simpler to work with is all

Answer 32

oh nw i get it thnq bro

Answer 33

good luck :)

Answer 34

:-)