## Monday, August 15, 2016

### கண்டுபிடி!

$$\frac{200^2+199^2+198^2+197^2+\dots+2^2+1^2}{200^2-199^2+198^2-197^2+\dots+2^2-1^2}$$

1. தொகுதி -பின்னத்தில் மேலே இருப்பது
1^2 + 2^2 + ... + 200^ = n (n+1)(2n+1)/6 = (200 x 201 x 401)/6 =2686700 ...... (1)
இப்பொழுது,
1^2 + 3 ^2 + ... 199^2 = n (2n+1) (2n-1)/3 = (100x201x199)/6 = 1333300 .... (2)

ஆக, பகுதி -பின்னத்தில் கீழே இருப்பது (1) - (2)= 1333400

ஆக, பின்னத்தின் மதிப்பு 2686700/1353400
=1.985 1485 1485 ...
சரியான விடை மாதிரித் தெரியவில்லை :)

சரவணன்

Saravanan

=

1. In the denominator, alternate signs are negative.

2. Yes, that's why I subtracted the sum of all odd terms [which is 1333300] from the sum of the whole series and got 1353400 for the denominator. For the numerator I have 2686700, hence got the fraction as 2686700/1353400 = 1.985 1485 1485...
Saravanan

2. 3. 4. Kouchik is right. Sudhakar, you are wrong with your answer. The formula you have posted which I have not approved is slightly faulty. I can see how you have worked out, but you need to check your n & 2n.

1. (2n+1)/3 if n is odd then -ve, is not it correct. n=200, (2*200+1) = 401/3=133.6667

2. (2n+1)/3 is correct. But you need to define "n" correctly. You wouldn't have gotten 200/3 as answer then.

5. I think the answer is 1.

6. 7. N = 200*201*401/6
x2 - (x-1)2 = 2x-1
D = 2(200+198+..+2) - 100

8. 9. (200*201*401/6)/((200*201*101/3)-(100*201*199/3))
=2686700/(1353400-1333300)
=2686700/20100
=133.666667

10. Numerator 200 X 201 X 401/6=2686700;
denominator (1+2)(1-2)+(3+4)(3-4)...(199+200)(199-200)=(-3-7-...-399) which is an Arithmatic progression =n(a1+an)/2=-100*(402)/2=-20100
So
2686700/-20100=-133.66