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Combinations — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsCombinations3 Marks
Question
For all positive integers n, show that ${^{2\text{n}}}\text{C}_{\text{n}}+{^{2\text{n}}}\text{C}_{\text{n}-1}=\frac{1}{2}\big({^{2\text{n+2}}}\text{C}_{\text{n}+1}\big).$
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Answer

We have,
$\text{L.H.S.}={^{2\text{n}}}\text{C}_{\text{n}}+{^{2\text{n}}}\text{C}_{\text{n}-1}$
$=\frac{2\text{n}!}{\text{n}!\text{n}!}+\frac{2\text{n}!}{(\text{n}-1)!(\text{n}-1)!}$
$=(2\text{n})!\Big[\frac{1}{\text{n}(\text{n}-1)!(\text{n})(\text{n}-1)!}+\frac{1}{(\text{n}-1)!(\text{n}-1)!}\Big]$
$=\frac{(2\text{n}!)}{(\text{n}-1)!(\text{n}-1)!}\Big[\frac{1+\text{n}^{2}}{\text{n}^{2}}\Big]\ ...(\text{i})$
${^{2\text{n+2}}}\text{C}_{\text{n}+1}=\frac{(2\text{n}+2)!}{(\text{n+1}!)(\text{n}+1)!}$
$=\frac{(2\text{n}+2)(2\text{n}+1)(2\text{n}!)}{\text{n}(\text{n}+1)\text{n}(\text{n}-1)!}\ ...(\text{ii})$
$=\frac{(2\text{n}!)}{(\text{n}-1)!(\text{n}-1)!}\times\frac{(\text{n}+1)^{2}(\text{n}^{2})(\text{n}-1)!(\text{n}-1)!}{(2\text{n}-2)(2\text{n}+1)(2\text{n}!)}\times\Big(\frac{1+\text{n}^{2}}{\text{n}^{2}}\Big)$
$=\frac{(\text{n}+1)(\text{n}^{2}+1)}{(2\text{n}+1)}\times\frac{1}{2}$

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