Sum of Product of Numbers taken two at a time

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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







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  • 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















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







share|cite|improve this question



















  • 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













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







share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Sep 2 at 12:22

























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













  • 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
















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