Reccurence

first 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}

1) or let b_n = a_n+1

So that the recurrence looks like b_{n+1} = 2b_n

ya 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

a_{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 {b_n} as b_n = a_n - \frac{2}{n^2} then this sequence obeys the recurrence b_{n+1} = \frac{b_n}{2}

Q4. na_n = (n-1)S_n
(n-1)a_n-1 = (n-2)S_n-1

Subtracting,
n(a_n - a_n-1) + a_n-1 = (n-1)a_n + S_n-1
=> a_n - (n-1)a_n-1 = (n-1)a_n-1/(n-2)

=> a_n = (n-1)²a_n-1/(n-2) = (n-1)²(n-2)(n-3)(n-4)....2.a₂
also 2a₂ = 1 + 1 + a₂
=> a₂ = 2

=> a_n = 2(n-1)(n-1)!

quite long i guess

5) 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}

thanx for replies prophet sir ,asish and akari

:)

btw q2 is left

continuing 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}}

ya 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 )

for 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)