Question

Sum of the series 1 + 2.2 + 3.2^2 + 4.2^3.....+ 100.2^99 is?

Answer 1

\[a_n=n\times2^{n-1}\]

Answer 2

gimmick is this is the derivative

Answer 3

The answer should be 99.2^100 + 1

Answer 4

i take it these are multiplication signs, not decimal points right?

Answer 5

Aha..u r point there @satellite73

Answer 6

\[1+r+r^2+r^2+...+r^n=\frac{r^{n+1}-1}{r-1}\]

Answer 7

take the derivative of both sides and evaluate at \(r=2\)

Answer 8

and it is AGP

Answer 9

@satellite73 @amistre64 @mukushla r u there

Answer 10

lol...everyone is here

Answer 11

Come...On......PLzz help

Answer 12

man satellite mentioned all the things u need to solve

Answer 13

i Did nt understand..

Answer 14

I think it is AGP..!

Answer 15

So.. it is Very Difficult to solve by that.....IF u have a easy method..plzz Tell me

Answer 16

oh its very easy man...can u take derivative?

Answer 17

Yup.. i knw to differentiate...

Answer 18

but did nt understand..satterlite

Answer 19

\[1+r+r^2+r^2+...+r^{100}=\frac{r^{101}-1}{r-1}\]take derivative of both sides u will get:
\[1+2r+3r^2+4r^2+...+100r^{99}=\frac{101r^{100}(r-1)-(r^{101}-1)}{(r-1)^2}\]now just let \(r=2\)

Answer 20

101.2^100 - 2^101 + 1

Answer 21

yeah and we are done

Answer 22

but should it be 101.2^100 - 2^100 + 1

Answer 23

yup..i got it,,,))

Answer 24

man be carefull...

\[101\times2^{100} - 2^{101} + 1=101\times2^{100} -2\times 2^{100} + 1=99\times2^{100}+ 1\]

Answer 25

thxx..... Can u tell me Why we r differen it.??

Answer 26

i only knw some basics on Calculus

Answer 27

because we want to change it to the form that we need

Answer 28

any else ...i think more eassy
Let P = 1 + 2.2 + 3.2^2 + 4.2^3.....+ 100.2^99
times 2 to both side, thus we get :
2P = 2 + 2.2^2 + 3.2^3 + 4.2^4.....+ 100.2^100
2P - P = - 1 - 2 - 2^2 - 2^3 - 2^4 - ... - 2^99 + 100*2^100
P = - (1+2+2^2+2^3+2^4+...+2^99) + 100*2^100
P = - (1(2^100 - 1)/(2-1)) +100*2^100
P = - 2^100 + 1 + 100*2^100
P = 99*2^100 + 1

Answer 29

thats really easier and better @RadEn
thank u

Answer 30

yw) but i need long time to solve it.... :D