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Question

Question: 1\(\angle 1 + 2\angle 2 + 3\angle 3 + ... + n\angle n\) is equal to...

1∠1+2∠2+3∠3+...+n∠n\angle 1 + 2\angle 2 + 3\angle 3 + ... + n\angle n is equal to

A

∠n+1‾\angle\underline{n + 1}

B

∠n+1‾\angle\underline{n + 1} - 1

C

∠n+1‾\angle\underline{n + 1} + 1

D

None

Answer

∠n+1‾\angle\underline{n + 1} - 1

Explanation

Solution

1∠1+2∠2+3∠3+...+n∠n1\angle 1 + 2\angle 2 + 3\angle 3 + ... + n\angle n

= ∑r=1nr∠r 6mu= 6mu∑r=1n{(r+1) 6mu− 6mu1} 6mu∠r\sum_{r = 1}^{n}{r\angle r\mspace{6mu} = \mspace{6mu}\sum_{r = 1}^{n}{\{(r + 1)\mspace{6mu} - \mspace{6mu} 1\}\mspace{6mu}\angle r}}

= ∑r=1n{(r+1) 6mu∠r 6mu− 6mu∠r}\sum_{r = 1}^{n}{\{(r + 1)\mspace{6mu}\angle r\mspace{6mu} - \mspace{6mu}\angle r\}}

= ∑r=1n(∠r+1‾−∠r)\sum_{r = 1}^{n}{(\angle\underline{r + 1} - \angle r)}.

= (∠2−∠1) 6mu+ 6mu(∠3−∠2)+(∠4−∠3)+...+(∠n+1‾−∠n)(\angle 2 - \angle 1)\mspace{6mu} + \mspace{6mu}(\angle 3 - \angle 2) + (\angle 4 - \angle 3) + ... + (\angle\underline{n + 1} - \angle n)

= (∠n+1‾ 6mu− 6mu∠1 6mu= 6mu∠n+1‾ 6mu− 6mu1(\angle\underline{n + 1}\mspace{6mu} - \mspace{6mu}\angle 1\mspace{6mu} = \mspace{6mu}\angle\underline{n + 1}\mspace{6mu} - \mspace{6mu} 1