Sum of Product of Numbers taken two at a time
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
So the question is
Find the Sum of Product taken two at a time of of numbers $1,2,2^2,2^3...2^n-2,2^n-1$
My approach was to create a double summation like this $y=sum_i=0^n-1sum_j=0^n-12^icdot2^j$ but the answer comes out to be wrong.
What's wrong with my approach?
Edit:
My answer comes out to be $(2^n-1-1)^2$ however my book mentions the answer as $frac132^2n-2^n-frac13$
calculus algebra-precalculus
asked Sep 2 at 12:13
Harshit Joshi
17512
add a comment |Â
up vote
1
down vote
favorite
So the question is
Find the Sum of Product taken two at a time of of numbers $1,2,2^2,2^3...2^n-2,2^n-1$
My approach was to create a double summation like this $y=sum_i=0^n-1sum_j=0^n-12^icdot2^j$ but the answer comes out to be wrong.
What's wrong with my approach?
Edit:
My answer comes out to be $(2^n-1-1)^2$ however my book mentions the answer as $frac132^2n-2^n-frac13$
calculus algebra-precalculus
asked Sep 2 at 12:13
Harshit Joshi
17512
2
You count things like $1times 2^1$ twice.
– lulu
Sep 2 at 12:21
Thanks for the quick answer
– Harshit Joshi
Sep 2 at 12:31
Yes. As this is an error I make all the time, it wasn't hard to spot.
– lulu
Sep 2 at 12:32
Easier to consider how the sum is changed when you add $2^n$ to the set, perhaps
– Mark Bennet
Sep 2 at 12:32
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
So the question is
Find the Sum of Product taken two at a time of of numbers $1,2,2^2,2^3...2^n-2,2^n-1$
My approach was to create a double summation like this $y=sum_i=0^n-1sum_j=0^n-12^icdot2^j$ but the answer comes out to be wrong.
What's wrong with my approach?
Edit:
My answer comes out to be $(2^n-1-1)^2$ however my book mentions the answer as $frac132^2n-2^n-frac13$
calculus algebra-precalculus
asked Sep 2 at 12:13
Harshit Joshi
17512
So the question is
Find the Sum of Product taken two at a time of of numbers $1,2,2^2,2^3...2^n-2,2^n-1$
My approach was to create a double summation like this $y=sum_i=0^n-1sum_j=0^n-12^icdot2^j$ but the answer comes out to be wrong.
What's wrong with my approach?
Edit:
My answer comes out to be $(2^n-1-1)^2$ however my book mentions the answer as $frac132^2n-2^n-frac13$
calculus algebra-precalculus
calculus algebra-precalculus
asked Sep 2 at 12:13
Harshit Joshi
17512
asked Sep 2 at 12:13
Harshit Joshi
17512
edited Sep 2 at 12:22
asked Sep 2 at 12:13
Harshit Joshi
17512
asked Sep 2 at 12:13
Harshit Joshi
17512
asked Sep 2 at 12:13
Harshit Joshi
17512
17512
2
You count things like $1times 2^1$ twice.
– lulu
Sep 2 at 12:21
Thanks for the quick answer
– Harshit Joshi
Sep 2 at 12:31
Yes. As this is an error I make all the time, it wasn't hard to spot.
– lulu
Sep 2 at 12:32
Easier to consider how the sum is changed when you add $2^n$ to the set, perhaps
– Mark Bennet
Sep 2 at 12:32
add a comment |Â
2
You count things like $1times 2^1$ twice.
– lulu
Sep 2 at 12:21
Thanks for the quick answer
– Harshit Joshi
Sep 2 at 12:31
Yes. As this is an error I make all the time, it wasn't hard to spot.
– lulu
Sep 2 at 12:32
Easier to consider how the sum is changed when you add $2^n$ to the set, perhaps
– Mark Bennet
Sep 2 at 12:32
2
2
You count things like $1times 2^1$ twice.
– lulu
Sep 2 at 12:21
You count things like $1times 2^1$ twice.
– lulu
Sep 2 at 12:21
Thanks for the quick answer
– Harshit Joshi
Sep 2 at 12:31
Thanks for the quick answer
– Harshit Joshi
Sep 2 at 12:31
Yes. As this is an error I make all the time, it wasn't hard to spot.
– lulu
Sep 2 at 12:32
Yes. As this is an error I make all the time, it wasn't hard to spot.
– lulu
Sep 2 at 12:32
Easier to consider how the sum is changed when you add $2^n$ to the set, perhaps
– Mark Bennet
Sep 2 at 12:32
Easier to consider how the sum is changed when you add $2^n$ to the set, perhaps
– Mark Bennet
Sep 2 at 12:32
add a comment |Â
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2902657%2fsum-of-product-of-numbers-taken-two-at-a-time%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
2
You count things like $1times 2^1$ twice.
– lulu
Sep 2 at 12:21
Thanks for the quick answer
– Harshit Joshi
Sep 2 at 12:31
Yes. As this is an error I make all the time, it wasn't hard to spot.
– lulu
Sep 2 at 12:32
Easier to consider how the sum is changed when you add $2^n$ to the set, perhaps
– Mark Bennet
Sep 2 at 12:32