Question

expanding (2k+1)^3 will get me (4k^3 + 12k^2 + 6k + 1)
from where we got the 4 and 12? following this http://en.wikipedia.org/wiki/Binomial_theorem I think i should get (2k^3 + 6k^2+6k+1) correct?

Answer 1

\[(2k+1)^3 \]
\[>> (4k^3+12k^2+6k+1)\]
from where we got the 4 and 12?

Answer 2

\[ (a+b)^3 = 3C0 \ a^3 b^0 + 3C1\ a^2 b^1 + 3C2\ a^1 b^2 + 3C3\ a^0 b^3 \]

where 3C0 means 3 choose 0 = 1, 3C1= 3, 3C2= 3 and 3C3=1

Answer 3

you should get \((2k)^3=8k^3\) for that first term.

Answer 4

in your problem a= 2k and b= 1

Answer 5

so you should get
\[ (2k + 1)^3 = 1 \cdot (2k)^3 \cdot1^0 + 3 \cdot (2k)^2\cdot 1^1 + 3 \cdot (2k)^1\cdot 1^2 + 1 \cdot (2k)^0 \cdot 1^3\]

Answer 6

of course that simplifies to
\[ 8k^3 +12 k^2 +6k +1 \]

Answer 7

Thanks.

Answer 8

yw