Find 3rd term of the expansion of (2 + 3x)^8

Question

anonymous · Answer

is there a way of doing this other than writing the expansion term by term?

anonymous · Answer

Is your expansion around x=0?

anonymous · Answer

If
$$ f(x)=(2 + 3x)^8\
f'(x) =24 (3 x+2)^7\
f''(x)=504 (3 x+2)^6\
\frac {f''(0)}{2!}=16128\
$$
The third term is:
$$16128 x^2
$$
If you expand f(x), you get
$$
f(x) = 6561 x^8+34992 x^7+81648 
   x^6\+108864 x^5+90720
   x^4+48384 x^3+16128 
   x^2\+3072 x+256
$$

amistre64 · Answer

you seem to have found the anti-3rd term :)

amistre64 · Answer

i spose that depends on how you define a proper polynomial eh ...

anonymous · Answer

the answer given in the book was 16128x^2 - i suppose it depends which way you order the terms  thanx guys

anonymous · Answer

The third term in a power series is of a functiom f at zero is
$$ \frac {f''(0)}{2!} x^2
$$
Since the series starts as
$$
\frac {f(0)}{0!} x^0+ \frac {f'(0)}{1!} x^1 + \frac {f''(0)}{2!} x^2+ \cdots
$$