I'm a complete beginner at these:

Could someone please correct me if I'm wrong please (and show me where)....

Here goes:

(3n+1)^3 = (3n)^3 + 3(3n)^2 + 3(3n)^2 + 1

which would give me

= 3 (9n^3 + 6n^2) +1 is this correct?

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- Feb 15th 2006, 12:07 AMNatashaBinomial Theorem for (3n+ 1)^3
I'm a complete beginner at these:

Could someone please correct me if I'm wrong please (and show me where)....

Here goes:

(3n+1)^3 = (3n)^3 + 3(3n)^2 + 3(3n)^2 + 1

which would give me

= 3 (9n^3 + 6n^2) +1 is this correct? - Feb 15th 2006, 04:04 AMDenMac21Quote:

Originally Posted by**Natasha**

$\displaystyle (3n+1)^3 = (3n)^3 + 3(3n)^2 + 3(3n)1^2 + 1$

$\displaystyle (3n+1)^3 = (3n)^3 + 3(3n)^2 + 3(3n) + 1$

I have marked with red ^2 in your post because in theorem its b^2 and because b=1 then b^2 =1, you have put (ab)^2 or a^2. - Feb 15th 2006, 01:26 PMNatasha
thanks :-)