Question

hmmm seems i understand mathematical induction now...but how to do it with sigma notations???

Answer 1

?? perhaps any example??

Answer 2

perhaps...here's a question from our last quiz that i totally have no idea what it means...

\[\LARGE \sum_{i = 1}^{n} 2^{i} = 2(2^{n} - 1)\]

any ideas @experimentX

Answer 3

it's a geometric progression

Answer 4

hmmm..so knowing that...how can this be solved?

Answer 5

http://www.mathisfunforum.com/viewtopic.php?id=10469

Answer 6

it's like
2^1 +2^1 +2^3 + ... +2^n = 2(2^n -1)

Answer 7

it's like
sum=2+2^2+2^3+2^4+....+2^2

Answer 8

lol.... it's 2^n in the last term :P

Answer 9

r x sum = 2^2+2^3+2^4+ ...+ 2^(n+1)

Answer 10

subtract = rx sum - sum

Answer 11

subtract = -2+(2^2 - 2^2) +(2^3 - 2^3) + ...+(2^n-2^n) + 2^(n+1) = 2^(n+1) - 2

Answer 12

subtract? isnt it going to be like substtute to n+1??

Answer 13

sum( r - 1) = 2(2^n-1)
sum = 2(2^n-1)/r => 4(2^n-1) ===> what the hell did i get???

Answer 14

gp = a(r^n-1)/(r-1)

Oops ... my fault, r=2, and r-1 = 1, so 2 instead of 4

Answer 15

im soooo confused T_T

Answer 16

take two Sn's
multiply one with 2 ---> let it be Sn2
find the value of Sn2-Sn => Sn(2-1) = Sn = that value.

Answer 17

wait...what's r??? and i think it's only proving so why take sums??

Answer 18

R = 2 common ratio.

Answer 19

ok thanks :DDD ill try comprehending it now...

Answer 20

yw