# Binomial expansion questions

• Mar 8th 2009, 07:43 AM
db5vry
Binomial expansion questions
I think I did this well but I'd like them checked so I know I have the right answers. Could you please look at these for me?

1] \$\displaystyle (1 + 2x)^6\$
= \$\displaystyle 1 + 12x + 60x^2 + 160x^3 + 240x^4 + 96x^5 + 64x^6\$

2] \$\displaystyle (3 + 2x)^5\$
= \$\displaystyle 243 + 810x + 1080x^2 + 730x^3 + 240x^4 + 32x^5\$

3] \$\displaystyle (5 + 3x)^3\$
= \$\displaystyle 125 + 225x + 135x^2 + 27x^3\$

4] \$\displaystyle (4 - 2x)^4\$
= \$\displaystyle 256 - 512x + 384x^2 - 108x^3 + 16x^4\$

Thanks if you can help with this!
• Mar 8th 2009, 08:08 AM
niranjan
Quote:

Originally Posted by db5vry
I think I did this well but I'd like them checked so I know I have the right answers. Could you please look at these for me?

1] \$\displaystyle (1 + 2x)^6\$
= \$\displaystyle 1 + 12x + 60x^2 + 160x^3 + 240x^4 + 96x^5 + 64x^6\$

2] \$\displaystyle (3 + 2x)^5\$
= \$\displaystyle 243 + 810x + 1080x^2 + 730x^3 + 240x^4 + 32x^5\$

3] \$\displaystyle (5 + 3x)^3\$
= \$\displaystyle 125 + 225x + 135x^2 + 27x^3\$

4] \$\displaystyle (4 - 2x)^4\$
= \$\displaystyle 256 - 512x + 384x^2 - 108x^3 + 16x^4\$

Thanks if you can help with this!

1, 2, 4 are not correct.
• Mar 8th 2009, 08:20 AM
db5vry
Quote:

Originally Posted by niranjan
1, 2, 4 are not correct.

Okay I made a mistake with a coefficient in the first one - I saw that and corrected it, but I worked out 2 and 4 again:

2] \$\displaystyle (3 + 2x)^5 \$
= \$\displaystyle 125 + 10x + 40x^2 + 80x^3 + 80x^4 + 32x^5 \$

4] \$\displaystyle (4 - 2x)^4 \$
= \$\displaystyle 256 - 8x + 24x^2 - 48x^3 + 16x^4 \$

Are these right now?
• Mar 8th 2009, 08:36 AM
niranjan
Quote:

Originally Posted by db5vry
Okay I made a mistake with a coefficient in the first one - I saw that and corrected it, but I worked out 2 and 4 again:

2] \$\displaystyle (3 + 2x)^5 \$
= \$\displaystyle 125 + 10x + 40x^2 + 80x^3 + 80x^4 + 32x^5 \$

4] \$\displaystyle (4 - 2x)^4 \$
= \$\displaystyle 256 - 8x + 24x^2 - 48x^3 + 16x^4 \$

Are these right now?

2] \$\displaystyle (3 + 2x)^5 \$
=\$\displaystyle 243 + 810x + 1080x^2 + 720x^3 + 240x^4 + 32x^5\$

4] \$\displaystyle (4 - 2x)^4 \$
= \$\displaystyle 256 - 512x + 384x^2 - 128x^3 + 16x^4 \$