The correct option is D 2^n (n0)+(n1)+(n2) +... (nn) (1+x)^n = ^nC_0 +^nC_1x + ^nC_2x^2+ ...+^nC_n+x^n Put x = 1 #160; ⇒ #160; 2^n = ^n ...

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Sum of n Terms

n0+ n1+ n2+...

Question

$(\frac{n}{0})$ + $(\frac{n}{1})$ + $(\frac{n}{2})$ +..........+ $(\frac{n}{n})$ =

2^{n - 1}

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^{2 n} C_{n}

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2^{n}

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2^{n + 1}

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Solution

The correct option is D $2^{n}$
$(\frac{n}{0}) + (\frac{n}{1}) + (\frac{n}{2}) + . . . - (\frac{n}{n})$

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

Put x = 1

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

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Similar questions

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

1. {C_{0}}^{2} + 3. {C_{1}}^{2} + 5. {C_{2}}^{2} + . . . + (2 n + 1) . {C_{n}}^{2} = 2 n .^{2 n - 1} C_{n} +^{2 n} C_{n} .

Prove that $^{2 n} C_{n} = \frac{2^{n} \times [1.3.5... (2 n - 1)]}{n!} .$

\frac{1^{2}}{(1) (3)} + \frac{2^{2}}{(3) (5)} + . . . . . + \frac{n^{2}}{(2 n - 1) (2 n + 1)} = \frac{n (n + 1)}{2 (2 n + 1)}

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1^{2} + {2.2}^{2} + 3^{2} + {2.4}^{2} + 5^{2} + {2.6}^{2} + \dots . i s \frac{n}{2} (n + 1)^{2}

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