Math, asked by manojkumar52595, 6 months ago

If p is a prime number, show that the coefficients of the terms
of $(1+x)^{p-1}$ are alternately greater and less by unity than some multiple
of p.

Answers

Answered by ADARSHBrainly

${\large{\bf{\pink{꧁. \: Given \: .꧂}}}}$

p is a prime number.

${\large{\bf{\pink{꧁. \: To \: show \: .꧂}}}}$

The coefficients of the terms
of ${\sf{(1+x)^{p-1}}}$ are alternately greater and less by unity than some multiple
of p.

${\Large{\bf{\red{꧁. \: Solution \: .꧂}}}}$

Let the coefficient be :-

${\sf{C_0 , \: C_1, \: C_2 --- C_r}}$

Then :-

$\\ { \bf{C_r= \frac{(p - 1)(p - 2)(p - 3) - - -(p - r)}{r ! } }}$

$\\ \implies{\bf{ \frac{ M(P)+(-1)r }{r! }}}$

$\\ \implies{\bf{ \frac{ M(P) }{r! }+(-1)^{r} }}$

${\bf{\underline{Now, C_r \: is \: a \: positive \: integer; \: hence}}}$

$\\ \implies{\sf{ \frac{ M(P) }{r! }+(-1)^{r} \: must \: be \: the \: positive \: integer}}$

and p is a prime number(given) ,then it must be the multiple of p.

Therefore:-

${\Large{\underline{\boxed{\color{blue}{ \bf{C_r = M(p) + (-1)^r}}}}}}$

Hence, Proved.

Answered by skmadhuri114

Step-by-step explanation:

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11th

Maths

Binomial Theorem

Special Cases of Binomial Theorem

If p is a prime number, sho...

MATHS

If p is a prime number, show that the coefficients of the terms of (1+x)

p−1

are alternately greater and less by unity than some multiple of p.

Hard

ANSWER

Let the co efficients be denoted by C

,.........C

,,..

then C

(p−1)(p−2)(p−3)...(p−r)

M(p)+(−1)r!

M(p)

+(−1)

Now, C

is a positive integer;hence

M(p)

+(1)

must be a positive integer, and since p is a prime number it must be a multiple of p; therefore

=M(p)+(−1)

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Hall & Knight Higher Algebra Chapter 23

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If p is a prime number, show that the coefficients of the terms of are alternately greater and less by unity than some multiple of p.

Answers

If p is a prime number, show that the coefficients of the terms
of $(1+x)^{p-1}$ are alternately greater and less by unity than some multiple
of p.