Sum manipulation

Question

my question is if these two sums are equal:

$\sum_{j=1}^n \lambda u_jx_j + \mu v_j x_j = \sum_{j=1}^n \lambda u_jx_j + \sum_{j=1}^n \mu v_jx_j$

Thank you for your help.

Answer 1

In general, we have

$$\sum_{i=1}^n(a_i+b_i)=\sum_{i=1}^na_i+\sum_{i=1}^nb_i$$

We can check it by using mathematical induction.

• Check base case

• Suppose it is true for $n-1$ terms.

\begin{align}\sum_{i=1}^n(a_i+b_i)&=\left(\sum_{i=1}^{n-1}(a_i+b_i)\right)+(a_n+b_n) \\ &=\left(\sum_{i=1}^{n-1}a_i+\sum_{i=1}^{n-1}b_i\right)+(a_n+b_n)\\ &=\sum_{i=1}^{n-1}a_i + \left( \left(\sum_{i=1}^{n-1}b_i\right)+(a_n+b_n)\right) \text{,associative}\\ &=\sum_{i=1}^{n-1}a_i + \left( \left(\sum_{i=1}^{n-1}b_i\right)+(b_n+a_n)\right) \text{,commutative}\\ &=\sum_{i=1}^{n-1}a_i + \left( \left(\sum_{i=1}^{n-1}b_i+b_n\right)+a_n\right) \text{,associative}\\ &=\sum_{i=1}^{n-1}a_i + \left( \sum_{i=1}^{n}b_i+a_n\right) \end{align}

I am leaving the last few steps for exercise.