Answers

2014-07-17T17:42:14+05:30
Sum from n=1 to 100 from 2^n=

n=1 => 2^1=2
n=2 => 2^2=4
n=3 => 2^3=8
n=4 => 2^4=16
...............
n=99=> 2^99
n=100 => 2^100

So
2+4+8+16+....+2^100  

now the tricky part :)

But we know that if we multiply the number with 1 the result will be the same
1xX=X doh... :)))

so we can say instead of 1: 2-1

So the ecuation will be:
(2-1)(2+4+8+16+....+2^99+2^100)=
(2-1)(2^1+2^2+2^3+....+2^99+2^100)=
 when we multiply we add the powers so 2^3 for example comes from 2*2^2=2^(2+1)=2^3


(2-1)(2^1+2^2+2^3+....+2^99+2^100)=

      2^2+2^3+2^4+...................+2^100+2^101 - 
2^1-2^2-2^3-2^4-..............-2^99-2^100

so remains 2^1+2^101=
2+2^101=
2(1+2^100)

0
where you don't understand tell me
no, question is _ 1*2^1+2*2^2+3*2^3+4*2^4+.....+100*2^100
pfff....so i was right...it is from n=0 to 100
sorry,but we put n=0,then .n*2^n=0
2014-07-17T21:11:59+05:30
See this can be done as
  S=1*2 + 2*2^2 + 3*2^3...........100*2^100
2S=         1*2^2 + 2*2^3...........  99*2^100 + 100*2^101
-              -        -                       -             -
-------------------------------------------------------------------------------------------------
-S=2 + 2^2 + 2^3 +............2^100 -100*2^101
using sum of gp formula we will get
-S=2(2^100-1) -100* 2^101 
S=100*2^101 - 2^101 - 2
S=2^101(99) - 2
0