Question

find the sum of the series

\(\large\tt \color{black}{1.2+2.2^2+3.2^3....100.2^{100}}\)

Answer 1

\(\Large\tt \color{black}{\sum\limits_{n=1}^{100}n.2^n}\) = ?

Answer 2

that is a huge sum

Answer 3

http://www.wolframalpha.com/input/?i=sum+n%3D1+to+100+n*2%5En

Answer 4

yes but there must be formula

Answer 5

i will try to think on it

Answer 6

@kirbykirby

Answer 7

no luck

Answer 8

\(\Large\tt \color{black}{\sum\limits_{n=1}^{100}n.2^n=(n-1)\times2^{101}+2}\)


see if it is correct

Answer 9

\(\Large\tt \color{black}{\sum\limits_{n=1}^{100}n.2^n=(99)\times2^{101}+2}\)

Answer 10

Proof by Induction:
For n=1 we have 1*2^1=2=(1-1)*2^(1+1)+2 true
Now assume for some integer k we have
\[\sum_{i=1}^{k}i 2^i=(k-1)2^{k+1}+2\]

\[\sum_{i=1}^{k+1}i2^i=(k+1)2^{k+1}+\sum_{i=1}^{k}i2^i \\= (k+1)2^{k+1}+(k-1)2^{k+1}+2 \\=2^{k+1}(k+1+k-1)+2\\=2^{k+1}(2k)+2 \\= k2^{k+2}+2\]
There for all integer n we have
\[\sum_{i=1}^{n}i2^i=(n-1)2^{n+1}+2\]

Answer 11

how did you come up with that

Answer 12

i used the options given ,but i still cant get it

a.) \(\Large\tt \color{black}{=(100)\times2^{101}+2}\)
b.) \(\Large\tt \color{black}{=(99)\times2^{100}+2}\)
c.) \(\Large\tt \color{black}{=(99)\times2^{101}+2}\)
d.) none

Answer 13

\[1\cdot2+2\cdot 2^2+3\cdot 2^3\]
say we wanted to look at this instead...
\[1\cdot 2+2\cdot 2^2+3 \cdot 2^3 \\ \\ \text{ I wonder if we can somehow } \\ \text { consider the geometric sequences below \to help} \\ 1 \cdot 2 +1 \cdot 2^2+ 1 \cdot 2^3=\frac{1-2^{3+1}}{1-2}=1(2^{3+1}-1) \\ 2 \cdot 2+ 2 \cdot 2^2+ 2 \cdot 2^3=2 \frac{1-2^{3+1}}{1-2}=2( 2^{3+1}-1) \\ 3 \cdot 2 +3 \cdot 2^2+3 \cdot 2^3=3\frac{1-2^{3+1}}{1-2}=3(2^{3+1}-1)\]

Answer 14

we have our terms that we care about occurring on that diagonal there

Answer 15

maybe we need to find some pattern for the numbers off the diagonal

Answer 16

so we can subtract them out easily

Answer 17

yes u r right

Answer 18

but is it possible

Answer 19

just thinking and writing my thoughts here
r1+2^4=r2

Answer 20

so maybe we don't need row 2

Answer 21

\[3(2^{3+1}+1)-1(2^{3+1}+1)\]

Answer 22

that should be equal to \[(3-1)2^{3+1}+2\]

Answer 23

ok yeah this is it

Answer 24

nope got lost again

Answer 25

but how can u say that we dont need second row

Answer 26

because we can extend the finite series in the first row to get all the terms in the second row

Answer 27

I will continue to think on this
i have to go
and that is just a whole bunch of me thinking
this problem is probably way easier than i'm making it

Answer 28

ok thanks very much

Answer 29

best I can come with his observing a pattern in the smaller series and formulate the conjecture (you can also choose to prove your conjecture if you are that level) it

or even easier just trying your little formulas to see which works

sorry i couldn't be more use

Answer 30

http://www.pagalguy.com/discussions/quant-by-arun-sharma-25023813/7176214

Answer 31

omg
that was way easier than anything i was thinking

Answer 32

hey but i didnt get that one can u explain that with the link

Answer 33

S = -(1.2 + 1.2^2 + ..............+2^100) + 100.2^101 ???

Answer 34

ohh ok that leaves(2-1) in bracket

Answer 35

u know in exams people get hardly 2 min per this types of questions with options

Answer 36

hey can that logic be used to derive formula too

Answer 37

pretend we have the little series we were talking about earlier
\[\text{ Let } S=1 \cdot 2 +2 \cdot 2^2+3\cdot 2^3 \\ \text{ then } 2S=1 \cdot (2 \cdot 2 )+2 \cdot (2^2 \cdot 2)+3 \cdot (2^3 \cdot 2) \\ \text{ so subtracting S from 2S will give us S, the \sum we are looking for} \\ 2S-S=1(2^2)+2(2^3)+3(2^4)-[1(2)+2(2^2)+3(2^3)] \\2S-S=-1(2)+(1-2)2^2+(2-3)2^3+3(2^4)\]

Answer 38

like they just put the "like terms together"
the parts that had the same exponent

Answer 39

\[2S-S=-1(2)+(-1)2^2+(-1)2^3+3(2^4) \\2S-S=(-1)[2+2^2+2^3]+3(2^4)\]

Answer 40

\[2S-S=(-1)(2)[1+2+2^2]+3(2^4) \\ 2S-S=(-1)(2)[2^{2}-1]+3(2^4)\\2S-S=(-1)(2)(2^2-1)+3(2^4) \\\]

Answer 41

ok i got that

Answer 42

\[S=-2(2^3)+2+3(2^4)=-2^4+3(2^4)+2=2\cdot 2^4+2\]

Answer 43

should be (2^3-1) above not (2^2-1)

Answer 44

ok thnks very much again

Answer 45

well I don't know how much help i was actually lol

Answer 46

lol me too