Question

2. Use the ideas in drawings (a) and (b) to find the solution to

Answer 1

Answer:

Step-by-step explanation:

Hello, I will compute twice the sum so I need to compute,

$\begin{aligned}1&+2&+...+&(n-1)&+n\\\\n&+(n-1)&+...+&2&+1\\\\-&-----&---&---&---\\\\(n+1)&+(n+1)&+...+&(n+1)&+(n+1)\end{aligned}$

1+n = n+1

2 + n-1=n+1

...

n-1+2=n+1

n+1=n+1

So, this is n times n+1 so it is n*(n+1)

$2(1+2+3...+n)=n(n+1)\\\\1+2+3...+n=\dfrac{n(n+1)}{2}$