1057if it is till n terms and 1 bracket is considered as a single term...
first of all first term in the bracket is same as n and no of terms is also n
so nth term=n2(2n+(n-1)1)
therefore it is n/2[3n-1]=3n22-n2
using summation formulas...
we get n2(n+1)2
1How to do this ?
1+(2+3)+(4+5+6)+,,,,upto n brackets,,,,
1057
Ketan Chandak
·2012-01-27 18:49:08
dis is simpler....
last term of the nth bracket is n(n+1)2
we alrdy knw dat sum of first a natural nos is a(a+1)2
jus replace a as n(n+1)2 in the above formula.... :)
