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Grade: 8


n greaterthanorequal to 2 And given that a1=1 How to approach such questions?

8 months ago

Answers : (3)

25122 Points
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.
hope you understand
8 months ago
Aditya Gupta
2068 Points
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.
8 months ago
Vikas TU
13786 Points
8 months ago
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