CameraIcon

SearchIcon

MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Polynomial and Its General Form

If P=n n2-1...

Question

If $P = n (n^{2} - 1^{2}) (n^{2} - 2^{2}) (n^{2} - 3^{2}) . . . (n^{2} - r^{2}), n > r, n ϵ N,$ then P is divisible by

A

(2 r + 2)!

No worries! We‘ve got your back. Try BYJU‘S free classes today!

B

(2 r - 1)!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

C

(2 r + 1)!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

D

none of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct options are
B $(2 r - 1)!$
D $(2 r + 1)!$
$P = n (n^{2} - 1) (n^{2} - 2^{2}) . . . (n^{2} - r^{2})$
$P = n (n - 1) (n + 1) (n - 2) (n + 2) . . . (n - r) (n + r)$
Rearranging the terms we get,
$P = (n - r) (n - r + 1) . . . (n + r - 1) (n + r)$
This is a product of $2 r + 1$ consecutive terms. Hence it must be divisible by $(2 r + 1)!$
Hence, options B and C are correct.

Suggest Corrections

thumbs-up

0

Similar questions

p

is a prime, show that

| p - 2 r - ----- - | 2 r - 1 - ----- - - 1

is divisible by

p

P = E I

E = I R, P =

I^{2} R

\frac{I}{2}

E^{2} R

\frac{E^{2}}{R}

S_{n} = n \sum r = 1 \frac{2 r + 1}{r^{4} + 2 r^{3} + r^{2}}

S_{10}

50 \sum r = 1 {tan}^{- 1} \frac{1}{2 r^{2}} = p,

then the value of

tan p

(\begin{matrix} p q \end{matrix})

p < q

q ϵ W

\infty \sum r = 0

(\begin{matrix} n 2 r \end{matrix})

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Polynomial and Its General Form

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Polynomial and Its General Form

Standard IX Mathematics

Join BYJU'S Learning Program

CrossIcon