Binomial

37th

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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

$(1+3^{1/3}+7^{1/7})^{10}=\displaystyle \sum _{r_{1}+r_{2}+r_{3}=10} \frac{10!}{r_{1}!r_{2}!(10-r_{1}-r_{2})!} 1^{(10-r_{1}-r_{2})}3^{(r_{1}/3)}7^{(r_{2}/7)}$

For a term to be free of radical $r_{1}$ and $r_{2}$ should be divisible by $3$ and $7$ respectively and also $r_{1}+r_{2}≤10$ .

So, possible pairs of $(r_{1},r_{2})=(0,0),(3,0),(6,0),(9,0),(0,7)$ and $(3,7)$ .

Akshat Sharda - 3 years, 5 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}`	$\boxed{123}$