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

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8 months ago

```							Dear student a1 = 1a2 = 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
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8 months ago
```							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 cc^2*a(n-2)= c^3*a(n-3) + c^2f(n-2).continuing this till 2, and then adding all the eqns, we getan = 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 :))
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8 months ago
```							Dear student Please refer the below link https://www.askiitians.com/forums/Algebra/n-greaterthanorequal-to-2-and-given-that-a1-1-how_259548.htmGood Luck
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8 months ago
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