Answer
False
Work Step by Step
since, we have $\sum_{i=1}^{2}a_ib_i=a_1b_1+a_2b_2$
and
$\sum_{i=1}^{2}a_i\sum_{i=1}^{2}b_i=(a_1+a_2)(b_1+b_2)$
or, $=a_1b_1+a_2b_2+a_1b_2+a_2b_1$
or, $=\sum_{i=1}^{2}a_ib_i+a_1b_2+a_2b_1$
or,$\sum_{i=1}^{2}a_ib_i=\sum_{i=1}^{2}a_i\sum_{i=1}^{2}b_i-(a_1b_2+a_2b_1)$
Hence, the given statement is false.
