If 1^2 +2^2 +3^2+.....100^2 = X And 1^2+2^2+3^2+........50^2 = Y Then 1^2-2^2+3^2-4^2+......+99^

M4maths Previous Puzzles

28 Jul 2010 Find the value

If 1^2 +2^2 +3^2+.....100^2 = X
And 1^2+2^2+3^2+........50^2 = Y
Then 1^2-2^2+3^2-4^2+......+99^2-100^2 = ?

OPtion

1) X-3Y
2) X-6Y
3) 4X-8Y
4) X-2Y
5) X-Y
6) 2X-Y
7) X-4Y
8) X-8Y
9) 3X-4Y
10) 4X-3Y

Solution

1^2-2^2+3^2-4^2+........+99^2-100^2
= (1^2 +2^2 +3^2+.....100^2)-2(2^2+4^2+6^2+8^2+……….+100^2)
= (1^2 +2^2 +3^2+.....100^2)-2.2^2(1^2+2^2+3^2+4^2+……….+50^2)
= X – 8 Y

Correct Option: 8 Best Solution (10)

Tridib Das   13 years AGO

1^2-2^2+3^2-4^2+......+99^2-100^2
= (1^2 +2^2 +3^2+.....100^2) - 2(2^2+4^2+6^2+.....+100^2)
= (1^2 +2^2 +3^2+.....100^2) - 2{2^2(1^2+2^2+3^2+........50^2)}
= X - 2{2^2(Y)}
= X - 8Y.

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SUDESHNA SEN   13 years AGO

1^2-2^2+3^2-4^2+......+99^2-100^2
=(1^2+3^2+5^2+......+99^2)- (2^2+4^2+6^2+......+100^2)
=(1^2 +2^2 +3^2+.....100^2)-(2^2+4^2+6^2+......+100^2)- (2^2+4^2+6^2+......+100^2)
=(1^2 +2^2 +3^2+.....100^2)-2^2 (1^2+2^2+3^2+........50^2)-2^2 (1^2+2^2+3^2+........50^2)
=X-4Y-4Y= X-8Y..... Option 8)

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ASHOK ROY   13 years AGO

1^2-2^2+3^2-4^2+......+99^2-100^2
=(1^2+3^2+5^2+......+99^2)- (2^2+4^2+6^2+......+100^2)
=(1^2 +2^2 +3^2+.....100^2)-(2^2+4^2+6^2+......+100^2)- (2^2+4^2+6^2+......+100^2)
=(1^2 +2^2 +3^2+.....100^2)-2^2 (1^2+2^2+3^2+........50^2)-2^2 (1^2+2^2+3^2+........50^2)
=X-4Y-4Y= X-8Y..... Option 8)

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sanjib biswas   13 years AGO

1^2-2^2+3^2-4^2+......+99^2-100^2
=(1^2+3^2+5^2+......+99^2)- (2^2+4^2+6^2+......+100^2)
=(1^2 +2^2 +3^2+.....100^2)-(2^2+4^2+6^2+......+100^2)- (2^2+4^2+6^2+......+100^2)
=(1^2 +2^2 +3^2+.....100^2)-2^2 (1^2+2^2+3^2+........50^2)-2^2 (1^2+2^2+3^2+........50^2)
=X-4Y-4Y= X-8Y..... Option 8)

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Prabir Sen   13 years AGO

1^2-2^2+3^2-4^2+......+99^2-100^2
=(1^2+3^2+5^2+......+99^2)- (2^2+4^2+6^2+......+100^2)
=(1^2 +2^2 +3^2+.....100^2)-(2^2+4^2+6^2+......+100^2)- (2^2+4^2+6^2+......+100^2)
=(1^2 +2^2 +3^2+.....100^2)-2^2 (1^2+2^2+3^2+........50^2)-2^2 (1^2+2^2+3^2+........50^2)
=X-4Y-4Y= X-8Y..... Option 8)

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Dr.Dipin Singh   13 years AGO

If 1^2 +2^2 +3^2+.....100^2 = X
And 1^2+2^2+3^2+........50^2 = Y

4*(1^2+2^2+3^2+........50^2) = 4Y
0r 2^2+4^2+....+100^2=4Y
2*(2^2+4^2+....+100^2)=8Y
Then 1^2-2^2+3^2-4^2+......+99^2-100^2 = X-8Y

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YadeeshKumar   13 years AGO

x=1^2+2^2+3^3+...+100^2
y=1^2+2^2+3^2+........50^2
therefore
x=(1^2+3^2+5^2+..+99^2)+2^2+4^2+..100^2
x=(1^2+3^2+5^2+..+99^2)+4(1^2+2^2+........50^2)
x=(1^2+3^2+5^2+..+99^2)+4y
x-4y=(1^2+3^2+5^2+..+99^2)--->A

1^2-2^2+3^2-4^2+......+99^2-100^2
=(1^2+3^2+5^2+..+99^2)-(2^2+4^2+..100^2)
=(1^2+3^2+5^2+..+99^2)-4(1^2+2^2+........50^2)
substitute A
=x-4y-4y
=x-8y
so ans is 8.

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A J V   13 years AGO

x=1^2+2^2+3^3+...+100^2
y=1^2+2^2+3^2+........50^2
since
x=(1^2+3^2+5^2+..+99^2)+4y
x-4y=(1^2+3^2+5^2+..+99^2)--->A
we need
1^2-2^2+3^2-4^2+......+99^2-100^2
=(1^2+3^2+5^2+..+99^2)-4(1^2+2^2+........50^2)
as per A
=x-4y-4y
=x-8y

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senthilkumaran   13 years AGO

Given, 1^2 +2^2 +3^2+.....100^2 = X
And 1^2+2^2+3^2+........50^2 = Y

1^2-2^2+3^2-4^2+......+99^2-100^2
= (1^2+3^2+..+99^2)-(2^2+4^2+..+100^2)
= [ (1^2+2^2+..+100^2)-(2^2+4^2+..+100^2)]-(2^2+4^2+..+100^2)
= [ x- (2^2+4^2+..+100^2)]-(2^2+4^2+..+100^2)
= [x - 2^2(1^2+2^2+..+50^2)] - 2^2(1^2+2^2+..+50^2)
= [x - 4y ] - 4y
= x-8y.

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Achelal Sharma   13 years AGO

The equation will come to
1^2 +2^2 +3^2+.....+100^2 - 2(2^2+4^2+6^2+...+100^2)
=> X-2(2^2(1^2 +2^2 +3^2+.....+50^2)
=>X-8Y

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