Dear student
 
a1 = 1
a2 = 2/3 + 4 – 15 = – 31/3
 
... similarly you can find rest terms. However, an is given then in what form, you want to ask.
 
Please tell
						

							
as usual, aruns answer is senseless, illogical and wrong.
correct method:
given an = c*a(n-1) + f(n), where c= 2/3 and f(n)= n^2 – 15.
similarly, replacing n by n – 1,
a(n-1)= c*a(n-2) + f(n-1)
or c*a(n-1)= c^2*a(n-2) + cf(n-1)
again, replacing by n – 2, and multiplying by c
c^2*a(n-2)= c^3*a(n-3) + c^2f(n-2).
continuing this till 2, and then adding all the eqns, we get
an = c^(n – 1)*a1 + ∑k=2 to n [(2/3)^(n – k)*f(k)]
= (2/3)^(n – 1)*1 + ∑k=2 to n [(2/3)^(n – k)*(k^2 – 15)]
now, we need to calculate S= ∑k=2 to n [(2/3)^(n – k)*(k^2 – 15)]
= ∑k=2 to n [(2/3)^(n – k)*k^2] – 15*∑k=2 to n [(2/3)^(n – k)]
clearly ∑k=2 to n [(2/3)^(n – k)] is a GP and hence the sum is easy to calculate.
the real difficulty lies in evaluating ∑k=2 to n [(2/3)^(n – k)*k^2]. since the calcs are too lengthy, i am only telling you the method:
we know that ∑k=0 to n x^k= S(x) (where S(x)= (1 – x^(n+1))/(1 – x) due to GP)
differntiate both sides wrt x
∑k=0 to n kx^(k – 1)= S’(x) 
or ∑k=0 to n kx^k= xS’(x) 
differentiate again
∑k=0 to n k^2x^(k – 1)= S’(x) + xS”(x)
∑k=0 to n k^2x^k= x(S’(x) + xS”(x))
so this is the basic method and in above prob, (2/3)^-1= 3/2 would have to be substituted as x.
it is an extremely lengthy ques which would never be asked in JEE, so no need to worry.
KINDLY APPROVE :))
 
						

							
Dear student 
Please refer the below link 
https://www.askiitians.com/forums/Algebra/n-greaterthanorequal-to-2-and-given-that-a1-1-how_259548.htm
Good Luck