- r -0 n C r 2 r - n 2-A- n -2 n -1B- n -2 nC- n -1 2 nD- n 2 2 n

The correct option is B n.2^n∑_r=0^n ^n C_r (2r n)^2 = ∑_r=0^n (4r^2 + n^2 4rn) ^nC_r = ∑_r=0^n 4r^2 ^nC_r + n^2 ∑_r=0^n ^nC_r 4n ∑_r=0^n ^n C_r = n. ...

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Byju's Answer

Standard XII

Mathematics

Binomial Coefficients

∑ r =0 n C r ...

Question

${n \sum r = 0}^{n} C_{r} (2 r - n)^{2} =$

n {.2}^{n + 1}

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n {.2}^{n}

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(n - 1) 2^{n}

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n^{2} 2^{n}

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Solution

The correct option is B $n {.2}^{n}$
$\sum_{r = 0}^{n}^{n} C_{r} (2 r - n)^{2} = \sum_{r = 0}^{n} (4 r^{2} + n^{2} - 4 r n)^{n} C_{r} = \sum_{r = 0}^{n} 4 r^{2}^{n} C_{r} + n^{2} \sum_{r = 0}^{n}^{n} C_{r} - 4 n \sum_{r = 0}^{n}^{n} C_{r} = n {.2}^{n}$

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Similar questions

Find the value of $\sum_{r = 0}^{n} r^{n} C_{r}$

Q. The coefficient of

x^{n}

in the polynomial

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

Q. If S_n =

\sum_{r = 1}^{n} \frac{1 + 2 + 2^{2} + . . . Sum to r terms}{2^{r}}

, then S_n is equal to
(a) 2ⁿ − n − 1

(b)

1 - \frac{1}{2^{n}}

(c)

n - 1 + \frac{1}{2^{n}}

(d) 2ⁿ − 1

n \sum r = 0 r^{2} . (^{n} C_{r})^{2}

is equal to

Q. There are non-square numbers between

n^{2}

and

(n + 2)^{2}

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n∑r=0nCr(2r−n)2=

The correct option is B n.2n∑nr=0nCr(2r−n)2=∑nr=0(4r2+n2−4rn)nCr=∑nr=04r2nCr+n2∑nr=0nCr−4n∑nr=0nCr=n.2n

${n \sum r = 0}^{n} C_{r} (2 r - n)^{2} =$

The correct option is B $n {.2}^{n}$
$\sum_{r = 0}^{n}^{n} C_{r} (2 r - n)^{2} = \sum_{r = 0}^{n} (4 r^{2} + n^{2} - 4 r n)^{n} C_{r} = \sum_{r = 0}^{n} 4 r^{2}^{n} C_{r} + n^{2} \sum_{r = 0}^{n}^{n} C_{r} - 4 n \sum_{r = 0}^{n}^{n} C_{r} = n {.2}^{n}$