If the number of positive integral solutions of x- 1-x- 2-x- 3-n be denoted by P- n then beginvmatrixPn - P-n-1- P-n-2- 1P-n-1- P-n-2- P-n-3P-n-2- P-n-3- P-n-4 vmatrix -

The correct option is C 1 ^nC_2 amp; ^n+1C_2 amp;^n+2C_2 ^n+1C_2 amp; ^n+2C_2 amp; ^n+3C_2 ^n+2C_2 amp; ^n+3C_2 amp; ^n+4C_2 = 1 ...

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Byju's Answer

Standard XII

Mathematics

Evaluation of a Determinant

If the number...

Question

If the number of positive integral solutions of x_1 +x_2+x_3=n be denoted by P_n then
$∣ ∣ ∣ ∣ \begin{matrix} P_{n} & P_{n + 1} & P_{n + 2} P_{n + 1} & P_{n + 2} & P_{n + 3} P_{n + 2} & P_{n + 3} & P_{n + 4} \end{matrix} ∣ ∣ ∣ ∣ =$

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Solution

The correct option is C
- 1

$∣ ∣ ∣ ∣ ∣ \begin{matrix} ^{n} C_{2} & ^{n + 1} C_{2} & ^{n + 2} C_{2}^{n + 1} C_{2} & ^{n + 2} C_{2} & ^{n + 3} C_{2}^{n + 2} C_{2} & ^{n + 3} C_{2} & ^{n + 4} C_{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = - 1$

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Similar questions

Q. If

n

and

p

be two

+

ive integers such that

n \geq p + 2

and

D (n, p) ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^{n} C_{p} & ^{n} C_{p + 1} & ^{n} C_{p + 2}^{n + 1} C_{p} & ^{n + 1} C_{p + 1} & ^{n + 1} C_{p + 2}^{n + 2} C_{p} & ^{n + 3} C_{p + 1} & ^{n + 2} C_{p + 2} \end{matrix} ∣ ∣ ∣ ∣ ∣

then prove that

D (n, p) = \frac{^{n + 2} C_{3}}{^{p + 2} C_{3}} D (n - 1, p - 1)

Let

p

be a prime number &

n

be a positive integer, then exponent of prime

p

n!

is denoted by

E_{p} (n!)

& is given by

E_{p} (n!) = [\frac{n}{p}] + [\frac{n}{p^{2}}] + [\frac{n}{p^{3}}] + . . . . . + [\frac{n}{p^{x}}]

where

x

is the largest positive integer such that

p^{x} \leq n < p^{x + 1}

and

[\cdot]

denotes the greatest integer

Again every natural number

N

can be expressed as the product of its prime factors given by

N = P_{1}^{k_{2}} P_{2}^{k_{2}} . . . . P_{r}^{k_{r}}

where

P_{1}, P_{2}, P_{3}, . . . . . . . P_{r}

are prime numbers &

k_{1}

are whole numbers.

The greatest integer

n

for which

77!

is divisible by

3^{n}

Q. Let

n

be a positive integer and define

f (n) = 1! + 2! + 3! + \dots + n!

.Find Polynomials

P (x)

and

Q (x)

such that

f (n + 2) = Q (n) f (n) + P (n) f (n + 1)

forall

n \geq 1

Find P(2).

Q. If

f (n) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} n & n + 1 & n + 2 {^{n} P}_{n} & {^{n + 1} P}_{n + 1} & {^{n + 2} P}_{n + 2} {^{n} C}_{n} & {^{n + 1} C}_{n + 1} & {^{n + 2} C}_{n + 2} \end{matrix} ∣ ∣ ∣ ∣ ∣

, where the symbols have their usual meanings. Then

f (n)

is divisible by

Q. If the value of determinant

∣ ∣ ∣ ∣ ∣ \begin{matrix} n^{2} & (n + 1)^{2} & (n + 2)^{2} (n + 1)^{2} & (n + 2)^{2} & (n + 3)^{2} (n + 2)^{2} & (n + 3)^{2} & (n + 4)^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = p \forall n \in N

and

p \in R .

Then the value of

p

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If the number of positive integral solutions of x_1 +x_2+x_3=n be denoted by P_n then ∣∣ ∣∣PnPn+1Pn+2Pn+1Pn+2Pn+3Pn+2Pn+3Pn+4∣∣ ∣∣=

The correct option is C - 1 ∣∣ ∣ ∣∣nC2n+1C2n+2C2n+1C2n+2C2n+3C2n+2C2n+3C2n+4C2∣∣ ∣ ∣∣=−1

If the number of positive integral solutions of x_1 +x_2+x_3=n be denoted by P_n then
$∣ ∣ ∣ ∣ \begin{matrix} P_{n} & P_{n + 1} & P_{n + 2} P_{n + 1} & P_{n + 2} & P_{n + 3} P_{n + 2} & P_{n + 3} & P_{n + 4} \end{matrix} ∣ ∣ ∣ ∣ =$

The correct option is C
- 1

$∣ ∣ ∣ ∣ ∣ \begin{matrix} ^{n} C_{2} & ^{n + 1} C_{2} & ^{n + 2} C_{2}^{n + 1} C_{2} & ^{n + 2} C_{2} & ^{n + 3} C_{2}^{n + 2} C_{2} & ^{n + 3} C_{2} & ^{n + 4} C_{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = - 1$