How will you solve?

Evaluate $\sum_{i=1}^{30} C_{i}^{2i}$

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

What's $C_i$ ? $i$ -th Catalan number?

Zi Song Yeoh - 8 years, 4 months ago

$C_{i}^{2i}$ denotes combinations.

Bakshinder Singh - 8 years, 4 months ago

same as ${2i \choose i}$

Bakshinder Singh - 8 years, 4 months ago

I thought it was i-th Catalan number to the power of 2i. :)

Zi Song Yeoh - 8 years, 4 months ago

It cant be solved by coefficent method approach. I tried!!!

Bakshinder Singh - 8 years, 4 months ago

Link

Harshit Kapur - 8 years, 4 months ago

Sorry, I didn't got your point! Here n=i and you can't take n as constant in summation.

Bakshinder Singh - 8 years, 4 months ago

i know, but it is good to know that theorum too :)

Harshit Kapur - 8 years, 4 months ago

The summation still looks ugly after using it. Anyway it is a good theorem to know about.

Yong See Foo - 8 years, 4 months ago

expand it as (iC0)^2 + (iC1)^2 +........ (iCi)^2 and then summate each term separately....

Jatin Yadav - 8 years, 3 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$