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Question

If {x} is the least integer greater than or equal to x, then find the value of the following series:
{1}+{2}+{3}+{4}+...+{99}+{100}
  1. 55
  2. 157
  3. 835
  4. 715


Solution

The correct option is D 715
Let S={1}+{2}+{3}+{4}+...+{99}+{100}S=1+2+2+2+3+3+3+3+3+...+10S=1+3(2)+5(3)+7(4)+9(5)+...+19(10)
Since Tn=(2n1)n=2n2nSn=(2n2n)=2n2 n
=2n(n+1)(2n+1)6n(n+1)2=n(n+1)2[23(2n+1)1]=n(n+1)2[4n+233]=n(n+1)(4n1)6S10=10×11×396=715

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