Find the value of $^{n + 1} C_{1}$ - $^{n + 1} C_{2}$ + $^{n + 1} C_{3}$ .................. $(- 1)^{n + 1}$ $^{n + 1} C_{n + 1}$

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Solution

Consider the expansion of $(1 + x)^{n + 1}$ ,

$(1 + x)^{n + 1}$ = ( $^{n + 1} C_{0}$ ) + ( $^{n + 1} C_{1}$ )x + ( $^{n + 1} C_{2}$ ) $x^{2}$ .................

Put x = -1

⇒ $(1 - 1)^{n + 1}$ = ( $^{n + 1} C_{0}$ ) - ( $^{n + 1} C_{1}$ ) + ( $^{n + 1} C_{2}$ ) - ( $^{n + 1} C_{3}$ ) ...............

⇒ ( $^{n + 1} C_{1}$ ) - ( $^{n + 1} C_{2}$ ) + ( $^{n + 1} C_{3}$ ).................. = $^{n + 1} C_{0}$

= 1

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Similar questions

\frac{1}{2}^{n} C_{1} .^{n - 1} C_{1} + \frac{2}{3}^{n} C_{2} + \frac{3}{4}^{n} C_{c} .^{n - 1} C_{3} + \dots \dots

If T= $^{2 n + 1} C_{1} +^{2 n + 1} C_{2} + . . . . . . +^{2 n + 1} C_{n} = 255$ . Find the value of n.

s_{n} = 1 + q + q^{2} + . . . + q^{n}

S_{n} = 1 + \frac{q + 1}{2} + {(\frac{q + 1}{2})}^{2} + . . . + {(\frac{q + 1}{2})}^{n}, q \neq 1

^{n + 1} C_{1} +^{n + 1} C_{2} . s_{1} +^{n + 1} C_{3} . s_{2} + . . . +^{n + 1} C_{n + 1} . s_{n} = k^{n} . S_{n}

^{2 n + 1} C_{0} +^{2 n + 1} C_{1} +^{2 n + 1} C_{2} + . . . . . . +^{2 n + 1} C_{n} +^{2 n + 1} C_{n + 1} +^{2 n + 1}

C_{n + 2} + . . . . . +^{2 n + 1} C_{2 n + 1}

n = 3

S_{n} = \sum_{k = 0}^{n} (- 1)^{K} .^{3 n} C_{k}, w h e r e n = 12, . . . . . .

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