Join now for JEE/NEET and also prepare for Boards Join now for JEE/NEET and also prepare for Boards. Register Now
Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
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?
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
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 :))
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
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -