induction proof manipulation problem
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starting with the LHS, how do you show
$(k/30)(k+1)(6k^3+9k^2+k-1)+(k+1)^4$= $(k+1)/30(k+2)(6(k+1)^3+9(k+1)^2+(k+1)-1$
as you can probably tell its part of my induction proof but i cant show this how is it done? thanks
Sorry edit its -1 on the end
proof-writing
asked Nov 20 at 1:55
hitherematey
577
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up vote
1
down vote
favorite
starting with the LHS, how do you show
$(k/30)(k+1)(6k^3+9k^2+k-1)+(k+1)^4$= $(k+1)/30(k+2)(6(k+1)^3+9(k+1)^2+(k+1)-1$
as you can probably tell its part of my induction proof but i cant show this how is it done? thanks
Sorry edit its -1 on the end
proof-writing
asked Nov 20 at 1:55
hitherematey
577
the main thing would be NOT to factor the quintic $ 6k^5 + 15k^4 + 10k^3 - k$ Since leaving off the denominator , add to that $30 (k+1)^4$
– Will Jagy
Nov 20 at 2:30
yep, that works
– Will Jagy
Nov 20 at 2:37
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
starting with the LHS, how do you show
$(k/30)(k+1)(6k^3+9k^2+k-1)+(k+1)^4$= $(k+1)/30(k+2)(6(k+1)^3+9(k+1)^2+(k+1)-1$
as you can probably tell its part of my induction proof but i cant show this how is it done? thanks
Sorry edit its -1 on the end
proof-writing
asked Nov 20 at 1:55
hitherematey
577
starting with the LHS, how do you show
$(k/30)(k+1)(6k^3+9k^2+k-1)+(k+1)^4$= $(k+1)/30(k+2)(6(k+1)^3+9(k+1)^2+(k+1)-1$
as you can probably tell its part of my induction proof but i cant show this how is it done? thanks
Sorry edit its -1 on the end
proof-writing
proof-writing
asked Nov 20 at 1:55
hitherematey
577
asked Nov 20 at 1:55
hitherematey
577
asked Nov 20 at 1:55
hitherematey
577
asked Nov 20 at 1:55
hitherematey
577
asked Nov 20 at 1:55
hitherematey
577
577
the main thing would be NOT to factor the quintic $ 6k^5 + 15k^4 + 10k^3 - k$ Since leaving off the denominator , add to that $30 (k+1)^4$
– Will Jagy
Nov 20 at 2:30
yep, that works
– Will Jagy
Nov 20 at 2:37
add a comment |
the main thing would be NOT to factor the quintic $ 6k^5 + 15k^4 + 10k^3 - k$ Since leaving off the denominator , add to that $30 (k+1)^4$
– Will Jagy
Nov 20 at 2:30
yep, that works
– Will Jagy
Nov 20 at 2:37
the main thing would be NOT to factor the quintic $ 6k^5 + 15k^4 + 10k^3 - k$ Since leaving off the denominator , add to that $30 (k+1)^4$
– Will Jagy
Nov 20 at 2:30
the main thing would be NOT to factor the quintic $ 6k^5 + 15k^4 + 10k^3 - k$ Since leaving off the denominator , add to that $30 (k+1)^4$
– Will Jagy
Nov 20 at 2:30
yep, that works
– Will Jagy
Nov 20 at 2:37
yep, that works
– Will Jagy
Nov 20 at 2:37
add a comment |
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the main thing would be NOT to factor the quintic $ 6k^5 + 15k^4 + 10k^3 - k$ Since leaving off the denominator , add to that $30 (k+1)^4$
– Will Jagy
Nov 20 at 2:30
yep, that works
– Will Jagy
Nov 20 at 2:37