Download App

Combinations — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSCombinations3 Marks
Question
For all positive integers n, show that ${^{2\text{n}}}\text{C}_{\text{n}}+{^{2\text{n}}}\text{C}_{\text{n}-1}=\frac{1}{2}\big({^{2\text{n+2}}}\text{C}_{\text{n}+1}\big).$
✓

Answer

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

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "(Each question 3 marks)" from Combinations→Combinations — all question types→MATHS STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

Find the sum of the following geometric series: 0.15 + 0.015 + 0.0015 + ... to 8 terms;
→
In a survey of 100 students, the number of students studying the various languages were found to be: English only 18, English but not Hindi 23, English and Sanskrit 8, English 26, Sanskrit 48, Sanskrit and Hindi 8, no language 24. Find:
  1. How many students were studying Hindi?
  2. How many students were studying English and Hindi?
→
Prove that $\cos\frac{\pi}{5}\cos\frac{2\pi}{5}\cos\frac{\pi}{5}\cos\frac{8\pi}{5}=\frac{-1}{16}$
→
Find $(a + b)^4 - (a - b)^4.$ Hence, evaluate ${(\sqrt 3 + \sqrt 2 )^4} - {(\sqrt 3 - \sqrt 2 )^4}$
→
If $a$ and $b$ are the roots $x^2-3 x+p=0$ and $c, d$ are roots of $x^2-12 x+q=0$ where $a, b, c, d$ form a G.P. Prove that $(q+p):(q-p)=$ 17:15.
→
Prove that: $\cos 6 x=32 \cos ^6 x-48 \cos ^4 x+18 \cos ^2 x-1$
→
If the sum of a certain number of terms of the A.P. $25, 22, 19$, ….. is $116$, find the last term.
→
Find the domain of the following real valued functions of real variable: $\text{f(x)}=\sqrt{\text{x}-2}$
→
Find the general solutions of the following equations: $\tan\text{mx}+\cot\text{nx}=0$
→
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{2\text{x}}{\sqrt{\text{a}+\text{x}}-\sqrt{\text{a}-\text{x}}}$
→