Question

Given a simple proof that$\sum _{k=0}^{n}3a_k=3\sum _{k=0}^n{a}_k$

Short answer

We have proved$\sum _{k=0}^{n}3a_k=3\sum _{k=0}^n{a}_k$

Step-by-step solution

Given information

We are given a relation$\sum _{k=0}^{n}3a_k=3\sum _{k=0}^n{a}_k$. We have to prove it

Proof

We have,

$$\begin{gathered} \sum _{k=0}^{n}3a_k \\ =3a_1+3a_2+....+3a_n \\ =3(a_1+a_2+....+a_n) \\ =3\sum _{k=0}^n{a}_k \\ =RHS \end{gathered}$$

Hence proved