^n{C_0} - frac{1}{2}{,^n}{C₁} + frac{1}{3}{,^n}{C₂} - ...... + {( - 1)^n}frac{{^n{Cₙ}}}{{n + 1

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7.Binomial Theorem

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$^n{C_0} - \frac{1}{2}{\,^n}{C_1} + \frac{1}{3}{\,^n}{C_2} - ...... + {( - 1)^n}\frac{{^n{C_n}}}{{n + 1}} = $

A

$n$

B

$1/n$

C

$\frac{1}{{n + 1}}$

D

$\frac{1}{{n - 1}}$

Solution

(c)Trick : Put $n = 1, 2$

At $n = 1,{\,^1}{C_0} – \frac{1}{2}{\,^1}{C_1} = 1 – \frac{1}{2} = \frac{1}{2}$

At $n = 2,\,{\,^2}{C_0} – \frac{1}{2}\,{\,^2}{C_1} + \frac{1}{3}{\,^2}{C_2} = 1 – 1 + \frac{1}{3} = \frac{1}{3}$

which is given by option $(c)$.

Standard 11

Mathematics

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