[แมวสามสี]

จาก $\frac{a_{n+2}}{a_n}$ = 2

$\sum_{n = 1}^{10}a_n= a_1+a_2+a_3+...+a_{10}$

31={$a_1+a_3+a_5+a_7+a_9$}+{$a_2+a_4+a_6+a_8+a_{10}$}

31={$a_1+2a_1+2^2a_1+2^3a_1+2^4a_1$}+{$a_2+2a_2+2^2a_2+2^3a_2+2^4a_2$}

31=$a_1(2^5-1)+a_2(2^5-1)$=$(a_1+a_2)(31)$

จะได้$ a_1+a_2=1$

$\sum_{n = 1}^{2552}a_n= a_1+a_2+a_3+...+a_{2552}$

={$a_1+a_3+a_5+...+a_{2551}$}+{$a_2+a_4+a_6+...+a_{2552}$}

={$a_1+2a_1+2^2a_1+...+2^{1275}a_1$}+{$a_2+2a_2+2^2a_2+...+2^{1275}a_2$}

=$a_1(2^{1276}-1)+a_2(2^{1276}-1)$=$(a_1+a_2)(2^{1276}-1)$

=$(2^{1276}-1)$