we have to find trace of matrix of order i , where 1<=i<=100.
so from the question we can say that ti follows a triplet where.
if ---> imod3=0 : di =2i+1 [email protected]
imod3=1 : di=i+3 [email protected]
i mod3=2 : di=0 [email protected]
for i = { 3,6,9................................99 } we use @1 (33 terms)
di={7,13,19.................................199} [email protected]
for i= {1,4,7..................................100} we use @2 (34 terms)
di={4,7,10,....................................103} [email protected]
for i={2,5,8.................................98} we use @3
di = {0,0,0,0,0,...............................} [email protected]
therefore trace of matrix = summation of di (1<=i<=100)
=using arithmetic progression sum eqn.
= (n/2) (first term +last term)
for eqn [email protected] :
di={7,13,19.................................199} : n=33, a=7 ,d=6, l=199
sum= n/2 (a+l) = 33/2 (199+7) =3399
for eqn @5 :
di={4,7,10,....................................103} : n=34 a=4 , d=3 ,l=103
sum = n/2 (a+l) = 34/2 (4+103) =1819
for eqn @6 :
di = {0,0,0,0,0,...............................}
sum =0
so total sum = 3399+1819+0 = 5218 answer.