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Question

If Sk=1+2+3+4+...+kk, find the value of S21+S22+S23+...+S2n.

Also, determine (Snn).


Solution

Given, Sk=1+2+3+...+kk=k(k+1)2k=k+12                  [ n=n(n+1)2]

   S21+S22+S23+...+S2n=nk=1S2k=nk=1(k+12)2=14nk=1(K+1)2=14nk=1(k2+2k+1)

= 14[nk=1k2+2nk=1k+nk=11]=14[n(n+1)(2n+1)6+2n(n+1)2+n]

= n4[(n+1)(2n+1)6+n+1+1]=n4(2n2+3n+1+6n+126)

= n24(2n2+9n+13)

Now,                  Snn=124(2n2+9n+13)=124(2n2+9n+131)

= 124[2×n(n+1)(2n+1)6+9×n(n+1)2+13n]

= n24[2×(2n2+3n+1)6+9n+92+13]

= n24(4n2+6n+2+27n+27+786)

= n144(4n2+33n+107)  


Mathematics

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