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Mathematics 18 Online
OpenStudy (anonymous):

4 ∑ (6k+1) K=1

OpenStudy (anonymous):

\[\sum_{r=1}^{n}(6r+1)=n+3n(n+1)\]here n=4 so ans=4+3*4(4+1)=64

myininaya (myininaya):

6*4(4+1)/2+1*4 =3*4*(5)+4 =60+4=64 yes right!

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