Question

Anyone knows summation properties??

Answer 1

The one with this \(\sum\)?

Answer 2

which ones would you like to know?

Answer 3

\[f(n)=\log_{2}(n)\]

Answer 4

then \[\sum_{k=2}^{100}f(n)\]

Answer 5

I believe that we have an arithmetic series here.

Answer 6

Let me wolfram it for a sec.

Answer 7

That 'k' would be 'n' right?

\[\Sigma \log_2n = \log_22 + \log_23 + ... + \log_2100 = \log_2(2\times 3 \times...\times 100) = \log_2(100!)\]


This is what I think, but I may be heading the wrong way totally.

Answer 8

it is f(2k)

Answer 9

\[\sum_{k=1}^{100}\log_2(2k)\]?

Answer 10

then it will be
1 + log(100!)

Answer 11

@TuringTest the upper one but k=2.

Answer 12

k=2 to 100.

Answer 13

\[\sum_{k=2}^{100}\log_2(2k)\]

Answer 14

\[\log(ab)=\log a+\log b\]

Answer 15

\[\sum_{k=2}^{100}\log_2(2k)=\sum_{k=2}^{100}\log_22+\log_2k=\sum_{k=2}^{100}1+\log_2(k)\]

Answer 16

then??????????

Answer 17

well, how many 1's are we gonna add?

Answer 18

99

Answer 19

so\[99+\sum_{k=2}^{100}\log_2k\]now what to to about log(k) exactly... I don't know any tricks offhand...

Answer 20

then \[\log_{2}(4*6*8*10*12...................................200)\]
??

Answer 21

we already took out the 2, so\[99+\log_2(2\cdot3\cdot3\cdots100)\]

Answer 22

uh oh, means I blooped.

Should be:
\[\sum_{k=2}^{100}\log_2(2k) = \log_2(2\times2)+ \log_2(2X3) +...+\log_2(2\times100)\]
\[=\log_2((2\times2\times2\times...\times2)(2\times3...100))\]
\[=\log_2(2^{100})+\log_2(100!)\]
\[=100\log_22 + \log_2(100!)\]
\[=100 + \log_2(100!)\]



This is better I guess.

Answer 23

oh sorry I made another mistake!! I am the grand daddy of stupidity >.<

Answer 24

\[99+\log_2(100!)\]is what I get

Answer 25

i've got options -
a)5010 c)5100
b)5050 d)5049

Answer 26

Yeah that would be a "99", am sorry I calculated one extra '2' there..

Answer 27

hm... now how do we be clever about \(\log_2(100!)\) is the question

Answer 28

wolfram doesn't give any of those answers

Answer 29

Yeah, I am not sure though that 100! would be a whole no. power of 2. And all the options are whole numbers, so.. :S

Answer 30

no, it's not a whole number

Answer 31

http://www.wolframalpha.com/input/?i=log_2%28100%21%29%2B99
so now let's play find the typo..

Answer 32

ans. is 5049.
i dont know how??

Answer 33

Well, then all the options set here are whole nos. and that makes me scratch my head.

If it's 5049, that means \(\log_2(100!)\) equals \(4950\) according to these folks (i.e. if we did it right).

Answer 34

I really don't see how \[\log_2(100!)\]can/should be a whole number
and as you can see above, wolf claims it to be about 525, way smaller than what we want

Answer 35

\[Suppose f(n)=\log_{2}(3)*\log_{3}(4)*\log_{4}5.........\log_{n-1}(n) \]
then the sum \[\sum_{k=2}^{100}=f(2^{k})\]

Answer 36

Sorry friends

Answer 37

it was f(2^k) not 2k.

Answer 38

ok, then\[\sum_{k=2}^{100}\log_2(2^k)\]really? this will be easy

Answer 39

oh no, the base is changing

Answer 40

oh dang, now this is a hard problem

Answer 41

@amistre64 challenging series question

Answer 42

Cant we take power of 2 out and solve it in an arithmetic progression??

Answer 43

how's the base changing?? :S

Answer 44

the way he wrote it above:
Suppose\[ f(n)=\log_{2}(3)*\log_{3}(4)*\log_{4}5.........\log_{n-1}(n)\]

Answer 45

@TuringTest when you simplify it it becomes log(base2)(n)

Answer 46

it does? o-0

Answer 47

\[f(n)=\log3/\log2 * \log4/\log * \log5/\log4 * ...........\log (n)/\log ()\]

Answer 48

We can observe that everything cancels out except log(n)/log2

Answer 49

ok, well then that's the hard part for me, but that is interesting. so then the series is an AP like you said.

Answer 50

http://www.wolframalpha.com/input/?i=2%2B3%2B4%2B5%2B...%2B100

Answer 51

ok thanks all.
I got it.