B Quickest way to calculate a given summation

[physsure]

How would you, personally, do this summation the quickest way?

 

[Charles Link]

It is an arithmetic sum: $S=(a+l)(\frac{n}{2})=(1+100)(100/2)$.

 

[WWGD]

How would you, personally, do this summation the quickest way?

You can use the idea by Gauss ( when he was 8 years or so) . Pair up:
1) 1+100=101
2) 2+99=101
...
...
50)50+51=101
There are 50 copies of 101 , for a total of 50(101)=5050.
WWGD: What would Gauss do? Probably same he did back then

 

[fresh_42]

How would you, personally, do this summation the quickest way?

$$50 \,\pi + \dfrac{25\cdot 67 \cdot 101}{\pi} - \dfrac{1}{2\pi}\sum_{k=1}^{100} (\pi-k)^2$$

 

[Charles Link]

$$50 \,\pi + \dfrac{25\cdot 67 \cdot 101}{\pi} + \dfrac{1}{2\pi}\sum_{k=1}^{100} (\pi-k)^2$$

@fresh_42 I believe your 3rd term needs a minus in front of it. $\\$ Note: $\sum\limits_{k=1}^{n} k^2=\frac{(2n+1)(n+1)(n)}{6}$ is the most difficult term to evaluate in the above. The rest is relatively straightforward.

 

[fresh_42]

@fresh_42 I believe your 3rd term needs a minus in front of it. $\\$ Note: $\sum\limits_{k=1}^{n} k^2=\frac{(2n+1)(n+1)(n)}{6}$ is the most difficult term to evaluate in the above. The rest is relatively straightforward.

Yep, thanks. Lost while turning pages in my scribble book.

 

[fresh_42]

The rest is relatively straightforward.

How about:
$$
\dfrac{\sqrt{5}^9\cdot 101}{2\,\pi^2} \cdot \sum_{n=1}^{\infty} \left( \dfrac{1}{n} \right)^2\cdot \dfrac{1}{n+1}\cdot \dfrac{F_n\cdot L_n}{C_n}
$$
with the Fibonacci sequence $F_n$, the Lucas sequence $L_n$, and the Catalan sequence $C_n$.

 

[Charles Link]

How about:
$$
\dfrac{\sqrt{5}^9\cdot 101}{2\,\pi^2} \cdot \sum_{n=1}^{\infty} \left( \dfrac{1}{n} \right)^2\cdot \dfrac{1}{n+1}\cdot \dfrac{F_n\cdot L_n}{C_n}
$$
with the Fibonacci sequence $F_n$, the Lucas sequence $L_n$, and the Catalan sequence $C_n$.

I do think @WWGD probably had the best answer for the OP in post 3.