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JEE Main 2015 · 10 Apr · Shift 1, Mathematics question 69

The value of ∑r=1630(r+2)(r−3)\displaystyle\sum_{r=16}^{30}(r+2)(r-3) is equal to :

Choose one option
Answer and worked solution

Official answer: B

∑r=1630(r2−r−6)\sum_{r=16}^{30}(r^2-r-6):

  • ∑r2=9455−1240=8215\sum r^2=9455-1240=8215;
  • ∑r=465−120=345\sum r=465-120=345;
  • ∑6=90\sum6=90.

8215−345−90=77808215-345-90=7780.

Topics:Special series sums
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