The term independent of x in 1-xn 1-1-x n is

The correct option is C C^2_0+C^1_1+C^2_2+ ......... +C^2_n (1+x)^n(1+1/x)^n = (C_0 +C_1x+C_2x^2 + ..... +C_nx^n) ( C_0 + C_1/x+C_2/x^2+ ....... + C_ ...

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Byju's Answer

Standard XII

Mathematics

Sum of Product of Binomial Coefficients

The term inde...

Question

The term independent of x in $(1 + x)^{n} {(1 + \frac{1}{x})}^{n}$ is

C_{0}^{2} + 2 \cdot C_{1}^{2} + 3 \cdot C_{2}^{2} + . . . . . . . . . . + (n + 1) \cdot C_{n}^{2}

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

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

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Solution

The correct option is C $C_{0}^{2} + C_{1}^{1} + C_{2}^{2} + . . . . . . . . . + C_{n}^{2}$
$(1 + x)^{n} (1 + 1 / x)^{n} = (C_{0} + C_{1} x + C_{2} x^{2} + . . . . . + C_{n} x^{n}) (C_{0} + \frac{C_{1}}{x} + \frac{C_{2}}{x^{2}} + . . . . . . . + \frac{C_{n}}{x^{n}})$
$∴$ The term independent of x is
$C_{0}^{2} + C_{1}^{2} + C_{2}^{2} + . . . . . . . . . . + C_{n}^{2}$

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Similar questions

Q. If

c_{0}, c_{1}, c_{2}, . . . . . . . c_{n}

denote the coefficients in the expansion of

{(1 + x)}^{n}

, prove that

{c_{0}}^{2} + {c_{1}}^{2} + {c_{2}}^{2} + . . . . {c_{n}}^{2} = \frac{| 2 n - - -}{| n - - | n - -}

Q. A.

^{2 n} C_{n} = C_{0}^{2} + C_{1}^{2} + C_{2}^{2} + C_{3}^{2} + \dots \dots + C_{n}^{2}

^{2 n} C_{n} =

term independent of

x

(1 + x)^{n} {(1 + \frac{1}{x})}^{n}

^{2 n} C_{n} = \frac{1.3.5.7 \dots \dots (2 n - 1)}{n!}

then

Q. If

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

, then

C_{0}^{2} + C_{1}^{2} + C_{2}^{2} + C_{3}^{2} + . . . . . . . + C_{n}^{2}

is equal to

Q. For

r = 0, 1, 2, . . . ., n

, prove that

C_{0} \cdot C_{r} + C_{1} \cdot C_{r + 1} + C_{2} \cdot C_{r + 2} + . . . . + C_{n - r} \cdot C_{n} =^{2 n} C_{(n + r)}

and hence deduce that
i)

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

ii)

C_{0} \cdot C_{1} + C_{1} \cdot C_{2} + C_{2} \cdot C_{3} + . . . . . + C_{n - 1} \cdot C_{n} =^{2 n} C_{n + 1}

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

$C_{0}^{2} + C_{1}^{2} + C_{2}^{2} + C_{3}^{2} + . . . . . . . . . . + C_{n}^{2}$ =

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The term independent of x in (1+x)n(1+1x)n is

The correct option is C C20+C11+C22+.........+C2n(1+x)n(1+1/x)n=(C0+C1x+C2x2+.....+Cnxn)(C0+C1x+C2x2+.......+Cnxn)∴ The term independent of x isC20+C21+C22+..........+C2n

The term independent of x in $(1 + x)^{n} {(1 + \frac{1}{x})}^{n}$ is