form 4 maths

The nth term of sequence of triangular numbers is given by :
Tn = n(n+1)/ 2

a) show that 231 is a triangular number.
Tn = n(n+1)/2
n(n+1)/2=231
n^2+n=462
n^2+n-462=0
(n-21)(n+22)=0
n=21 or n=-22(rejected)
so 231 is the 21st term of the sequence of triangular numbers
____________________
b) . express T( n-1) + Tn in terms of n.
T(n-1) = (n-1)(n-1+1)/2=(n-1)(n)/2
Tn = n(n+1)/2
Therefore,
T(n-1) +T(n)
=(n-1)(n)/2+n(n+1)/2
=(n^2-n)/2+(n^2+n)/2
=((n^2-n)+(n^2+n))/2
=2n^2/2
=n^2
Sorry, The above person made a mistake in this park,you can compare my answer with his answer to check that
_____________________
As T(n-1) +T(n) means the sum of two consecutive numbers
so n^2 also means the sum of two consecutive numbers
n^2=484
n=22
So the two consecutive number are:
T(n-1)
=T(22-1)
=T(21)
= 21(21+1)/2
=231
T(n)
=T(22)
=22(22+1)/2
=253
SO The two numbers are 231 and 253

The nth term of sequence of triangular numbers is given by :
Tn = n(n+1)/2

a) . show that 231 is a triangular number.
b) . express T( n-1) + Tn in terns of n.
d). Hence find two consecutive triangular numbers such that their sum is 484
a)
when n=21
21*22/2=231
so 231 is a triangular number
b)
T( n-1) + Tn
=n(n+1)/2+(n+1)(n+2)/2
=1/2(n+1)(n+n+2)
=(n+1)^2
c)
484=22^2
let n=21
then
T(21) and T(22) is the required number
T(21)=231
T(22)=253