Let f n - begin vmatrix n - n-1 - n-2 -n P - n - n -1 P - n -1- n -1 P - n -2-n Cn -n-1 C-n-1-n-1 C-n-2lend vmatrix Then- f n is divisible byA- n2-n-1B- n-1 -C- In-D- none of these

The correct options are A n^2+n+1 C n! f(n)= n amp;n+1 amp;n+2 ^nP_n amp;^n+1P_n+1 amp;^n+2P_n+2 ^nC_n amp;^n+1C_n+1 amp;^n+2C_n+2 =n! n ...

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Byju's Answer

Standard XII

Mathematics

Evaluation of a Determinant

Let f n = beg...

Question

Let $f (n) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} n & n + 1 & n + 2^{n} P_{n} & ^{n + 1} P_{n + 1} & ^{n + 1} P_{n + 2}^{n} C_{n} & ^{n + 1} C_{n + 1} & ^{n + 1} C_{n + 2} \end{matrix} ∣ ∣ ∣ ∣ ∣$
Then, f(n) is divisible by

$n^{2} + n + 1$

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(n+1)!

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none of these

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Solution

The correct options are
A
$n^{2} + n + 1$

C
n!

$f (n) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} n & n + 1 & n + 2^{n} P_{n} & ^{n + 1} P_{n + 1} & ^{n + 2} P_{n + 2}^{n} C_{n} & ^{n + 1} C_{n + 1} & ^{n + 2} C_{n + 2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = n! ∣ ∣ ∣ ∣ ∣ \begin{matrix} n & (n + 1) & (n + 2) 1 & (n + 1) & (n + 1) (n + 2) 1 & 1 & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ Δ = n! ∣ ∣ ∣ ∣ ∣ \begin{matrix} n & 1 & 1 1 & n & (n + 1)^{2} 1 & 0 & 0 \end{matrix} ∣ ∣ ∣ ∣ ∣ (C_{3} \to C_{3} - C_{2}, C_{2} \to C_{2} - C_{1}) Δ = n! \times [(n + 1)^{2} - n] = n! \times (n^{2} + n + 1)$

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Similar questions

Q. f(n)={n/2,3n+1

Q. The value of the determinant

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1^{n} C_{n - 1} & ^{n + 1} C_{n} & ^{n + 2} C_{1}^{n} C_{n - 2} & ^{n + 1} C_{n - 1} & ^{n + 2} C_{2} \end{matrix} ∣ ∣ ∣ ∣ ∣

Let $f (n) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} n & n + 1 & n + 2^{n} P_{n} & ^{n + 1} P_{n + 1} & ^{n + 1} P_{n + 2}^{n} C_{n} & ^{n + 1} C_{n + 1} & ^{n + 1} C_{n + 2} \end{matrix} ∣ ∣ ∣ ∣ ∣$
Then, f(n) is divisible by

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. For a fixed positive integer n, if

D = ∣ ∣ ∣ ∣ ∣ \begin{matrix} n! & (n + 1)! & (n + 2)! (n + 1)! & (n + 2)! & (n + 3)! (n + 2)! & (n + 3)! & (n + 4)! \end{matrix} ∣ ∣ ∣ ∣ ∣

, then

[\frac{D}{(n!)^{3}} - 4]

is___

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Let f(n)=∣∣ ∣ ∣∣nn+1n+2nPnn+1Pn+1n+1Pn+2nCnn+1Cn+1n+1Cn+2∣∣ ∣ ∣∣ Then, f(n) is divisible by

The correct options are A n2+n+1 C n! f(n)=∣∣ ∣ ∣∣nn+1n+2nPnn+1Pn+1n+2Pn+2nCnn+1Cn+1n+2Cn+2∣∣ ∣ ∣∣=n!∣∣ ∣ ∣∣n(n+1)(n+2)1(n+1)(n+1)(n+2)111∣∣ ∣ ∣∣Δ=n!∣∣ ∣ ∣∣n111n(n+1)2100∣∣ ∣ ∣∣(C3→C3−C2,C2→C2−C1)Δ=n!×[(n+1)2−n]=n!×(n2+n+1)

Let $f (n) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} n & n + 1 & n + 2^{n} P_{n} & ^{n + 1} P_{n + 1} & ^{n + 1} P_{n + 2}^{n} C_{n} & ^{n + 1} C_{n + 1} & ^{n + 1} C_{n + 2} \end{matrix} ∣ ∣ ∣ ∣ ∣$
Then, f(n) is divisible by