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Question

If nr=1tr=nk=1kj=1ji=1(2), then nr=11tr is
  1. n+1n
  2. n+1
  3. nn+1
  4. 1n+1


Solution

The correct option is C nn+1
nr=1tr=nk=1kj=1ji=1(2)=nk=1kj=12j=nk=1k(1+k)tr=r(1+r)1tr=1r(r+1)=1r1r+1
Now nr=11tr
=1112+1213+1314+        +1n1n+1=11n+1=nn+1

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