1)find the nth term of the sequence
such that 
2)Find the nth term of the sequence such that
3)Find the nth term of the sequence such that
then Calculate 
4)Find the n th term of the sequence such that 
5)Find the n th term of the sequence such that 
1first one
a_n=3.2^{n-1}-1
\texttt{the characteristic polynomial here is } 1-2x \\ hence \\ a_n=coefficient (x^n,(1-2x)^{-1}\left(a_1x +x^2 +x^3+....x^n+.\infty \right) \\ hence \\ a_n =1+2+2^2+2^3+.......+2^{n-2}+a_1.2^{n-1} \\ \Rightarrow a_n=2^{n-1}-1+2^{n-1}.2=\boxed{3.2^{n-1}-1}
13rd one
a_n=\sum_{k=1}^{n}{a_k}-\sum_{k=1}^{n-1}{a_k} \\ a_n=3n(n+1)\\ \frac{1}{a_n}=\frac{1}{3}(\frac{1}{n}-\frac{1}{n+1})
3411) or let b_n = a_n+1
So that the recurrence looks like b_{n+1} = 2b_n
1ya akari u r correct :)
try oder ones also ....i added 5th one
btw i had an approch for 1st one ....but much lengthier than generating polynomials
341a_{n+1} = \frac{a_n}{2} + \frac{n^2-2n-1}{n^2(n+1)^2} = \frac{a_n}{2} + \frac{2}{(n+1)^2} - \frac{1}{n^2}
\Rightarrow a_{n+1} - \frac{2}{(n+1)^2}= \frac{1}{2} \left(a_n - \frac{2}{n^2} \right)
So if you define a sequence {bn} as b_n = a_n - \frac{2}{n^2} then this sequence obeys the recurrence b_{n+1} = \frac{b_n}{2}
106Q4. nan = (n-1)Sn
(n-1)an-1 = (n-2)Sn-1
Subtracting,
n(an - an-1) + an-1 = (n-1)an + Sn-1
=> an - (n-1)an-1 = (n-1)an-1/(n-2)
=> an = (n-1)2an-1/(n-2) = (n-1)2(n-2)(n-3)(n-4)....2.a2
also 2a2 = 1 + 1 + a2
=> a2 = 2
=> an = 2(n-1)(n-1)!
quite long i guess
3415) Rewriting the recurrence as n^3 a_n = (n-1)^3 a_{n-1} + n(n-1)
Let b_n = n^3 a_n
Then we have the recurrence b_n -b_{n-1} = n(n-1)
This can be summed up telescopically to arrive at
b_n-b_1 = \sum_{k=2}^n k^2-k = \sum_{k=1}^n k^2-k = \frac{n(n^2-1)}{3}
b1=0 and hence a_n = \frac{(n^2-1)}{3n^2}
23continuing from prophet sir's # 10 post
b_{n}=2b_{n+1}
b_{n-1}=2b_{n}
b_{n-2}=2b_{n-1}=4b_{n}=2^{2}b_{n}
\Rightarrow b_{n-k}=2^{k}b_{n}
\Rightarrow b_{n-(n-1)}=2^{(n-1)}b_{n}
\Rightarrow b_{1}=2^{(n-1)}b_{n}
\Rightarrow b_{n}= \frac{b_{1}}{2^{(n-1)} }
also\;\;\; b_{n}= a_{n}-\frac{2}{n^{2}}
b_{1}= a_{1}-\frac{2}{1}= -1
a_{n}-\frac{2}{n^{2}}= \frac{-1}{2^{n-1}}
a_{n}=\frac{2}{n^{2}}+\frac{-1}{2^{n-1}}= \frac{2^{n}-n^{2}}{2^{(n-1)}n^{2}}
1ya qwerty thats correct :)
btw u can do this
\\\texttt{after \; u\; got\; this}\\ a_{n+1}-\frac{2}{(n+1)^{2}}={1\over 2}\left(a_{n}-\frac{2}{n^{2}}\right)\\ \Rightarrow \prod_{k=1}^{k=n-1}\left ( a_{k+1}-\frac{2}{(k+1)^{2}} \right )=\prod_{k=1}^{k=n-1}\left ( {1\over 2}\left(a_{k}-\frac{2}{k^{2}}\right) \right )\\\\\Rightarrow a_{n}-\frac{2}{n^2}=\frac{1}{2^{n-1}}\left ( a_{1}-\frac{2}{{1}^{2}} \right )
1for 1st one oder way
a_{n+1}=2a_{n}+1\Longleftrightarrow a_{n+1}+1=2(a_{n}+1)
\therefore a_{n}+1=2^{n-1}(a_{1}+1)=3\cdot 2^{n-1},
threfore giving a_{n}=3\cdot 2^{n-1}-1\ (n\geq 1)
