Let n be a positive integer and, (1+x) n=a0+a1x+a2x2+⋯+anxn. What is a0+a1+a2+⋯+an equal to?

Given expansion is (1+x)n=a0+a1x+a2x2+....anxnSubstituting x=1 on both the sides we get,(1+1)n=a0+a1×1+a2×12+....an×1n⟹2n=a0+a1+a2+...anThus, the value of a0+a1+a2+...an is 2n.

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Let n be a positive integer and, { ( 1+x ) }^{ n }={ a }_{ 0

Question

Let $n$ be a positive integer and, $(1 + x)^{n} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{n} x^{n}$ . What is $a_{0} + a_{1} + a_{2} + \dots + a_{n}$ equal to?

1

2^{n}

2^{n - 1}

2^{n + 1}

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Michael Discussed

Let

n

be a positive integer and,

(1 + x)^{n} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{n} x^{n}

. What is

a_{0} + a_{1} + a_{2} + \dots + a_{n}

equal to?

12 mins ago

Discuss this question LIVE

12 mins ago

Text solutionVerified

Given expansion is

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

Substituting

x = 1

on both the sides we get,

(1 + 1)^{n} = a_{0} + a_{1} \times 1 + a_{2} \times 1^{2} + .... a_{n} \times 1^{n}

⟹ 2^{n} = a_{0} + a_{1} + a_{2} + ... a_{n}

Thus, the value of

a_{0} + a_{1} + a_{2} + ... a_{n}

2^{n}

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Question Text	Let $n$ be a positive integer and, $(1 + x)^{n} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{n} x^{n}$ . What is $a_{0} + a_{1} + a_{2} + \dots + a_{n}$ equal to?
Answer Type	Text solution:1
Upvotes	150

Let n be a positive integer and, (1+x) n=a0​+a1​x+a2​x2+⋯+an​xn. What is a0​+a1​+a2​+⋯+an​ equal to?

Text solutionVerified

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Let $n$ be a positive integer and, $(1 + x)^{n} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{n} x^{n}$ . What is $a_{0} + a_{1} + a_{2} + \dots + a_{n}$ equal to?