Question

(1-9/4)(1-9/25)(1-9/64)......(1-9/400)
Does anyone know how to solve this kind of problems?

Answer 1

First try to find the law of the numbers' change. In the numbers above, we can find that 4, 25, 64 are 2^2, 5^2, 8^2.

Answer 2

Yes i got that but i can never create an equation to solve the problem

Answer 3

Give me a minute please.

Answer 4

All right. We can find that 2+3=5, 5+3=8, so that denominator of next number is 8+3=11, then 14, 17, 20, and 20^2=400.

Answer 5

yes but is there a way quick way to solve these

Answer 6

Let me see.

Answer 7

OK. The denominators can be written as (2+0*3)^2, (2+1*3)^2, (2+2*3)^2, (2+3*3)^2, (2+4*3)^2, (2+5*3)^2, (2+6*3)^2.
(2+0*3)^2
=(2+0*3)(2+0*3)
=2*2+2(0*3)+2(0*3)+(0*3)*(0*3)
=2*2+(4+0*3)(0*3)
So there are two 2, and 2*2+0*3 0*3.
(2+1*3)^2
=(2+1*3)(2+1*3)
=2*2+2(1*3)+2(1*3)+(1*3)(1*3)
=2*2+(4+1*3)(1*3)

Thus we get this formula.
(2+3x)^2=4+(4+3x)(3x)

Answer 8

@Mathstudent97

Answer 9

i think it has something to do with

Answer 10

Yes. 3 is important.

Answer 11

In 4, x=0;
in 21,x=1;
in 64,x=2;
then x=3, 4, 5, 6.
The formula is (2+3x)^2=4+(4+3x)(3x).
(4+3x)(3x)=12x+9x^2
Now first find how many 12 there, then how many 9 there.

Answer 12

As x=0, 1, 2, 3, 4, 5, 6, so there are (0+1+2+3+4+5+6) 12.

Answer 13

(0+1+2+3+4+5+6)*12
=21*12
=252

Answer 14

thanks but i think I"l have to ask my teacher

Answer 15

I have not completed yet.

Answer 16

And there are (0^2+1^2+2^2+3^2+4^2+5^2+6^2+7^2) 9.
140*9=1260
And there are 7 numbers, so 7*(1-9)=7*(-8)=-56
the answer is -56/252+1260=
-28/756
=-27

Answer 17

@Mathstudent97