MHB Find Sum of first 100 Terms in $(n-2)a_n - (n-1)a_{n-1} +1=0$ with $a_{100}=199$

[Albert1]

$a_1,a_2,-----,a_n,--a_{100}\in R$
giving :
$(n-2)a_n - (n-1)a_{n-1} +1=0 $
and :$a_{100}=199$
here $(2\leq n \leq 100)$
find $a_1+a_2+a_3+------+a_{100}=?$

 

[chisigma]

$a_1,a_2,-----,a_n,--a_{100}\in R$
giving :
$(n-2)a_n - (n-1)a_{n-1} +1=0 $
and :$a_{100}=199$
here $(2\leq n \leq 100)$
find $a_1+a_2+a_3+------+a_{100}=?$

Just a question ... the difference equation that defines the $a_{n}$ is...

$\displaystyle a_{n}= \frac{(n - 1)\ a_{n- 1} - 1}{n - 2}\ (1)$

... isn't it?... now if we suppose to know $a_{1}$, what is the value of $a_{2}$ obtained from (1)?(Tmi)...

Kind regards

$\chi$ $\sigma$

 

[topsquark]

If you are going to go that route then we would have to let [math]n \to 100 - n[/math] as we only know [math]a_{100}[/math].

-Dan

 

[Albert1]

Just a question ... the difference equation that defines the $a_{n}$ is...

$\displaystyle a_{n}= \frac{(n - 1)\ a_{n- 1} - 1}{n - 2}\ (1)$

... isn't it?... now if we suppose to know $a_{1}$, what is the value of $a_{2}$ obtained from (1)?(Tmi)...

Kind regards

$\chi$ $\sigma$

If you define $a_n \,\, as \,(1)$ then $n\neq 2$

$a_2$ will be undefined

 

[chisigma]

If you define $a_n \,\, as \,(1)$ then $n\neq 2$

$a_2$ will be undefined

The problems is that if $a_{2}$ is undefined and You use (1) to find $a_{3}$, then $a_{3}$ is also undefined and the same is for $a_{4}$, $a_{5}$, ..., $a_{100}$...

Kind regards

$\chi$ $\sigma$

 

[Opalg]

$a_1,a_2,-----,a_n,--a_{100}\in R$
giving :
$(n-2)a_n - (n-1)a_{n-1} +1=0 $
and :$a_{100}=199$
here $(2\leq n \leq 100)$
find $a_1+a_2+a_3+------+a_{100}=?$

[sp]Claim: For $n\geqslant2$, $a_n = (n-1)a_2 - (n-2)$.

Proof by induction: For $n=2$ this is a tautology, $a_2=a_2$.

Suppose the result is true for $n$. Then the recurrence relation $(n-2)a_n - (n-1)a_{n-1} +1=0 $ tells us that $$(n-1)a_{n+1} = na_n - 1 = n(n-1)a_2 - n(n-2) - 1 = n(n-1)a_2 - (n-1)^2.$$ Therefore $a_{n+1} = na_2 - (n-1)$, which is what is required to prove the inductive step.

It then follows that $199 = a_{100} = 99a_2 - 98$. So $99a_2 = 297$, from which $a_2 = 3$.

Thus the formula for $a_n$ becomes $a_n = 3(n-1) - (n-2) = 2n-1$, for $n\geqslant2$. The same formula also holds for $n=1$, as you can see by putting $n=1$ in the given recurrence relation.

Hence $$\sum_{n=1}^{100} a_n = \sum_{n=1}^{100}(2n-1) = 100(101) - 100 = 100^2.$$[/sp]

 

[Albert1]

[sp]Claim: For $n\geqslant2$, $a_n = (n-1)a_2 - (n-2)$.

Proof by induction: For $n=2$ this is a tautology, $a_2=a_2$.

Suppose the result is true for $n$. Then the recurrence relation $(n-2)a_n - (n-1)a_{n-1} +1=0 $ tells us that $$(n-1)a_{n+1} = na_n - 1 = n(n-1)a_2 - n(n-2) - 1 = n(n-1)a_2 - (n-1)^2.$$ Therefore $a_{n+1} = na_2 - (n-1)$, which is what is required to prove the inductive step.

It then follows that $199 = a_{100} = 99a_2 - 98$. So $99a_2 = 297$, from which $a_2 = 3$.

Thus the formula for $a_n$ becomes $a_n = 3(n-1) - (n-2) = 2n-1$, for $n\geqslant2$. The same formula also holds for $n=1$, as you can see by putting $n=1$ in the given recurrence relation.

Hence $$\sum_{n=1}^{100} a_n = \sum_{n=1}^{100}(2n-1) = 100(101) - 100 = 100^2.$$[/sp]

nice solution!

 

[Albert1]

$a_1,a_2,-----,a_n,--a_{100}\in R$
giving :
$(n-2)a_n - (n-1)a_{n-1} +1=0 ---(1)$
and :$a_{100}=199$
here $(2\leq n \leq 100)$
find $a_1+a_2+a_3+------+a_{100}=?$

Spoiler

n=2, we get $a_1=1-----(2)$
n=3 :$a_3-2a_2=-1$------(3)
n=4 :$2a_4-3a_3=-1$------(4)
n=5 :$3a_5-4a_4=-1$------(5)
-----------------------------
-----------------------------
n=100 :$98a_{100}-99a_{99}=-1---(100)$
(3)+(4)+-------+(100):
$98a_{100}-2(a_2+a_3+---+a_{99})=-98$
$a_2+a_3------+a_{99}=9800$
$\therefore a_1+a_2+------+a_{100}=1+9800+199=10000$
in fact there is no need to find the value of $a_2$