Below given are two formulae for long summation. 1) 1 + 2 + 3 + 4 +......................... +n = n(n+1) / 2 2) 1 ^ 2 + 2 ^ 2 + 3 ^ 2 + 4 ^ 2 + ............................+ n ^ 2 = 1 / 6 n (n+1) (2n+1) What will be the value of 1% of 2 + 2% of 3 + 3% of 4 +...............+ 99% of 100 = ?
Answer is 3333
1% of 2 + 2% of 3 + 3% of 4 + ....... 99% of 100 = (1% of 1 + 2% of 2 + 3% of 3 + ....... 99% of 99) + (1% of 1 + 2% of 1 + 3% of 1 + ...... 99% of 1) = [(1^2 + 2^2 + ..... + 99^2)/100] + [(1+2+3+..... + 99)*1/100] = [1/6*(99)(100)(199)/100] + [1/2*(99)(100)(1)/100] = 3333
99! 8! 6! 5! 4!/100! 7! 3! 0! = ??
x, y, z and k are four non zero positive integers satisfying 1/x + y/2 = z/3 + 4/k, minimum integral value of k for integral value of x, y and z will be
If x, y, z are 3 non zero positive integers such that x+y+z = 8 and xy+yz+zx = 20, then What would be minimum possible value of x*y^2*z^2
If 1^2 + 2^2 + 3^2 + ................. 100^2 = A & 1^3 + 2^3 + 3^3 + .................. 100^3 = B Then 1^2*2 + 2^2*3 + 3^2*4 + 4^2*5 ........ 99^2*100 + 100^2*101=??
Which of the following numbers has the lowest value if: 1>2, 3<4, 6>7, 2>5, 7>3 and 5>4
