show that the general term of the sequence is Tn=n^2+3n

Question

please help me

Answer 1

a) Terms of quadratic sequence is of the form an^2 + bn +c

when n=1, i.e the first term = a+b+c =4 (given).  ------------------(1)

And the first difference is 3a+b = 6 (given) -----------------------(2)

and the second difference is 2a = 2 (given)

So a=1

Putting the value of a in eq(2) we get:

b=3

Putting the value of b in eq(1) we get:

c=0

So the general term, Tn = 1n^2 + 3n + 0 = n^2 +3n

b)

T1 = 4 (given)

T2 = 2^2 + 3*2 = 10

T3 = 3^2 + 3*3 = 18

c) T100 = 100^2 + 3*100 = 10000+300 = 10300