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Grade: 8

                        

Given that a1=1 and find an=_______ How can I approach such questions?

Given that 
a1=1   and find an=_______
How can I approach such questions?

11 months ago

Answers : (3)

Arun
25768 Points
							
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
11 months ago
Aditya Gupta
2075 Points
							
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 :))
 
11 months ago
Vikas TU
14148 Points
11 months ago
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