If C r -25 C r and C 0-5 - C 1-9 - C 2-101 - C 25-225- k k is equal to

S = ^25C_0 + 5 ^25C_1 + 9 ^25C_2 + ⋯ + 97 ^25C_24 + 101 ^25C_25 = 2^25k ⋯(1 ) Reverse and apply property ^nC_r = ^nC_n r in all coefficients nbsp; S = 101 ...

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Byju's Answer

Standard XII

Mathematics

nCr Definitions and Properties

If C r =25 C ...

Question

If $C_{r} =^{25} C_{r}$ and $C_{0} + 5 \cdot C_{1} + 9 \cdot C_{2} + \dots + 101 \cdot C_{25} = 2^{25} \cdot k$ $k$ is equal to

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Solution

$S =^{25} C_{0} + 5^{25} C_{1} + 9^{25} C_{2} + \dots + 97^{25} C_{24} + 101^{25} C_{25} = 2^{25} k \dots (1)$
Reverse and apply property $^{n} C_{r} =^{n} C_{n - r}$ in all coefficients
$S = 101^{25} C_{0} + 97^{25} C_{1} + \dots + 5^{25} C_{24} +^{25} C_{25} \dots (2)$
Adding (1) and (2), we get
$2 S = 102 [^{25} C_{0} + 25 C_{1} + \dots +^{25} C_{25}]$
$S = 51 \times 2^{25}$
$\Rightarrow k = 51$

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Similar questions

Q. If

C_{r} =^{25} C_{r}

and

C_{0} + 5 \cdot C_{1} + 9 \cdot C_{2} + \dots + 101 \cdot C_{25} = 2^{25} \cdot k

k

is equal to

Q. If

C_{r}

stands for

{^{n} C}_{r}

then

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

is equal to

Q. The sum of the series

\frac{1}{1 \times 2} {^{25} C}_{0} + \frac{1}{2 \times 3} {^{25} C}_{1} + \frac{1}{3 \times 4} {^{25} C}_{2} + . . . \frac{1}{26 \times 27} {^{25} C}_{25}

\frac{C_{0}}{1} + \frac{C_{1}}{2} + \frac{C_{2}}{3} + . . . . . . . . . . + \frac{C_{100}}{101}

equals ?

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}

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If Cr= 25Cr and C0+5⋅C1+9⋅C2+⋯+101⋅C25=225⋅k k is equal to

S= 25C0+5 25C1+9 25C2+⋯+97 25C24+101 25C25=225k⋯(1) Reverse and apply property nCr= nCn−r in all coefficients S=101 25C0+97 25C1+⋯+5 25C24+25C25⋯(2) Adding (1) and (2), we get 2S=102[ 25C0+ 25C1+⋯+ 25C25] S=51×225 ⇒ k=51

If $C_{r} =^{25} C_{r}$ and $C_{0} + 5 \cdot C_{1} + 9 \cdot C_{2} + \dots + 101 \cdot C_{25} = 2^{25} \cdot k$ $k$ is equal to