If 1-1-x-1-x2-1-x3-1-xn-K-0n aK XK- then which of the following is true-A- a K - n C KB- a 92- a 82- n -1 C 10 n -1 C 10- n -1 C 8C- an-2-nn-1-2D- - K -0 n a K -2 n -1-1

The correct option is D ∑_K=0^na_K = 2^n+1 11+(1+x) + (1+x)^2 + (1+x)^3 + …… +(1+x)^n = ∑_K=0^na_K X^K So the coefficient of x^K in (1+x)^n+1 1/x=a_K=^n+1C_K+ ...

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Byju's Answer

Standard XII

Mathematics

Sum of Coefficients of All Terms

If 1+1+x+1+x2...

Question

If $1 + (1 + x) + (1 + x)^{2} + (1 + x)^{3} + \dots \dots + (1 + x)^{n} = \sum_{K = 0}^{n} a_{K} X^{K}$ , then which of the following is true?

a_{n - 2} = \frac{n (n - 1)}{2}

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a_{9}^{2} - a_{8}^{2} =^{n + 1} C_{10} (^{n + 1} C_{10} -^{n + 1} C_{8})

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a_{K} =^{n} C_{K}

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\sum_{K = 0}^{n} a_{K} = 2^{n + 1} - 1

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Solution

The correct option is D $\sum_{K = 0}^{n} a_{K} = 2^{n + 1} - 1$
$1 + (1 + x) + (1 + x)^{2} + (1 + x)^{3} + \dots \dots + (1 + x)^{n} = \sum_{K = 0}^{n} a_{K} X^{K}$
So the coefficient of $x^{K}$ in $\frac{(1 + x)^{n + 1} - 1}{x} = a_{K} =^{n + 1} C_{K + 1}$
$a_{n - 2} =^{n + 1} C_{n - 1} = \frac{n (n + 1)}{2}$
$a_{9}^{2} - a_{8}^{2} =^{n + 2} C_{10} (^{n + 1} C_{10} -^{n + 1} C_{9})$
$\sum_{K = 0}^{n} a_{K} =^{n + 1} C_{1} +^{n + 1} C_{2} + \dots \dots +^{n + 1} C_{n + 1} = 2^{n + 1} - 1$

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Similar questions

If $1 + (1 + x) + (1 + x)^{2} + (1 + x)^{3} + . . . . . . + (1 + x)^{n} = \sum_{k = 0}^{n} a_{k} x^{k},$ then which of the following is true?

Q. If x₁,x₂,x₃,x₄….x_2n+1are in Arithmetic Progression, then find the value of [(x_2n+1–x₁)/ (x_2n+1+x₁)] + [(x_2n–x₂)/ (x_2n–x₂)]………….. [(x_n+2–x_n)/ (x_n+2–x_n)]

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If 1+(1+x)+(1+x)2+(1+x)3+……+(1+x)n=∑nK=0aKXK, then which of the following is true?

The correct option is D ∑nK=0aK=2n+1−11+(1+x)+(1+x)2+(1+x)3+……+(1+x)n=∑nK=0aKXK So the coefficient of xK in (1+x)n+1−1x=aK=n+1CK+1 an−2=n+1Cn−1=n(n+1)2 a29−a28=n+2C10(n+1C10−n+1C9) ∑nK=0aK=n+1C1+n+1C2+……+n+1Cn+1=2n+1−1

If $1 + (1 + x) + (1 + x)^{2} + (1 + x)^{3} + \dots \dots + (1 + x)^{n} = \sum_{K = 0}^{n} a_{K} X^{K}$ , then which of the following is true?