Math, asked by saryka, 30 days ago

Q7. Show that 1¹⁹⁹⁷ + 2¹⁹⁹⁷ + ... + 1996¹⁹⁹⁷ is divisible by 1997.​

Answers

Answered by user0888
105

Short memo:-

The second solution is deleted as there is a mistake.

Solution A:-

Step 1. Polynomial identity

x^{n}+y^{n}=(x+y)(x^{n-1}-x^{n-2}y+...-xy^{n-2}+y^{n-1})

Step 2. Grouping the terms

(1^{1997}+1996^{1997})+(2^{1997}+1995^{1997})+...+(998^{1997}+999^{1997})

=(1+1996)(...)+(2+1995)(...)+...+(998+999)(...)

=1997(...)

Hence, the given number is divisible by 1997.

More information:-

Why does x^{n}+y^{n}=(x+y)(x^{n-1}-x^{n-2}y+...-xy^{n-2}+y^{n-1})?

Let's prove this right now. This is an identity for an odd exponent n. Let's try synthetic division. (Refer to the attachment.)

Attachments:
Answered by shiza7
61

Answer:

According to fermat's Little Theorem, Since 1997 is a prime number,

1¹+ ,2¹⁹⁹⁷... ,+ 1996 ¹

are divided by1997 then,remainder will be respectively 1,2,..,1996

1+2+..+1996 =

 \frac{1}{2 }  \times 1996 \times 1997

= 998 × 1997

Which is divisible by 1997.

Hence remainder =0

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