Calcule:
a) $1678^{2}-1677^{2}$
b) $1001^{2}+1000^{2}$
c) $19999^{2}$
d) $2001^{2}+2002^{2}+2003^{2}$
Fatorando temos:

$1678^{2}-1677^{2}=(1678+1677)(1678-1677)=3355$.

(b) Como $(a+b)^{2}=a^{2}+2 a b+b^{2}$, temos:

$$
\begin{aligned}
1001^{2}+1000^{2} & =(1000+1)^{2}+1000^{2}=1000^{2}+2000+1+1000^{2}= \\
& =2 \times 1000^{2}+2001=2002001
\end{aligned}
$$

(c) Como $(a-b)^{2}=a^{2}-2 a b-b^{2}$, temos:

$$
\begin{aligned}
19999^{2} & =(20000-1)^{2}=\left(2 \times 10^{4}\right)^{2}-4 \times 10^{4}+1= \\
& =4 \times 10^{8}-4 \times 10^{4}+1=399960001
\end{aligned}
$$

(d) Colocando em função de 2000 temos:

$$
\begin{aligned}
2001^{2}+2002^{2}+2003^{2} & =(2000+1)^{2}+(2000+2)^{2}+(2000+3)^{2}= \\
& =3 \times 2000^{2}+12 \times 2000+14 \\
& =12024014
\end{aligned}
$$