MATHS...

1.
a) T(1)=1=1^2
T(2)=4=2^2
T(3)=9=3^3
T(4)=16=4^4
T(5)=25=5^5
therefore, the general term=T(n)= n^2

b) T(1)=1/2=1/(1+1^1)
T(2)=1/6=1/(2+2^2)
T(3)=1/12=1/(3+3^3)
T(4)=1/20=1/(4+4^4)
T(5)=1/30=1/(5+5^5)
T(6)=1/42=1/(6+6^6)
therefore, the general term=T(n)=1/(n+n^n)

2.
a)T(1)=1+1^1=2
T92)=2+2^2=6
T(3)=3+3^3=12
therefore, the general term=T(n)=n+n^n
b) sum of first two term =T(1)+T(2)
=2+6
=8
c) let the term be n
1/(n+n^n)=110
n(n+1)=110=10*11=10*(10+1)
therefore, the 10th term has a value of 110

1a) n^2
b) 1/n(n+1)

2. 睇唔到d亂碼
''each term is equal to the sum of the square of the term number and the term number'',

1)
a) general term = n^2
b) 1 / [n*(n+1)]

2) I cannot read your statement clearly, "... and the term number ''...", what is '&#?