a)
The 11th and 12th term can be given be using n=11 and n=12
#-> 1/2*11*(11+1)+1/2*12*(12+1)#
#=1/2*11*12+1/2*12*13#
#=66+78 = 144=12^2#
Thus, the sum of the 11th and the 12th terms gives #12^2# so it is a perfect square.
b)
To find the #(n+1)th# term we can replace the #n# with #n+1# and see the result:
#1/2(n+1)((n+1)+1)=1/2(n+1)(n+2)#
As expected
I believe the additional #n# in the question may have been a typing error?
c) To prove this statement we must consider the #nth# and #(n+1)th# term demonstrated in part (b). #n# and #n+1# terms are consecutive integers so putting them into the triangular number formula and summing gets:
#1/2n(n+1) +1/2(n+1)(n+2)#
#=1/2(n(n+1)+(n+1)(n+2))#
Expand the brackets to get:
#1/2(n^2+n+n^2+3n+2)=1/2(2n^2+4n+2)#
At this point we can simplify the value by factoring out 2 and cancelling the half to get:
#n^2+2n+1=(n+1)^2# A square number, thus the statement for question c is proved.
