The sum ∑ r = 1 10 r 2 + 1 × r ! , is equal to:

The sum ∑r=110r2+1×r!, is equal to:
  1. 11 ×11!
  2. 10 ×11! 
  3. 11!
  4. 101 ×10!

Solution

∑r=110r2+1r!
Let us express the rth term, as a difference involving consecutive factorials, Tr=r2+1+r-r r! =r2+r r!-r-1r!
Therefore,
⇒Tr=rr+1!-r-1r! 
Putting r=1,2,3,...,10 and adding, we get ∑r=110rr+1!-r-1r! =10×11!
∴∑r=110r2+1×r!=10×11!

Asked in: JEE Main 2016 (10 Apr Online)

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