edited by
28,197 views
66 votes
66 votes
Consider the recurrence relation $a_1 =8 , a_n =6n^2 +2n+a_{n-1}$. Let  $a_{99}=K\times 10^4$. The value of $K$ is __________.
edited by

10 Answers

Best answer
204 votes
204 votes
$a_n=6n^2+2n+a_{n-1}$

$=6n^2+2n+6(n-1)^2+2(n-1)+a_{n-2}$

$=6n^2+2n+6(n-1)^2+2(n-1)+6(n-2)^2+2(n-2)+......+a_1$

$=6n^2+2n+6(n-1)^2+2(n-1)+6(n-2)^2+2(n-2)+......+6.1^2+2.1$

$=6(n^2+(n-1)^2+...+2^2+1^2)+2(n+(n-1)+...+2+1)$

$=6\times \frac{n(n+1)(2n+1)}{6} \;+ \; 2 \times\frac{n(n+1)}{2}$

$=n(n+1)(2n+1+1)$

$a_n=2n(n+1)^2$

for $n=99 \; \;a_{99}=2\times 99 \times(99+1)^2 =198 \times 10^4$
selected by
55 votes
55 votes
$a_n=6n^2+2n+a_{n-1}$

Solution = Homogeneous Solution + Particular Solution .................$(1)$

$Homogeneous\ Solution$,
$a_n=a_{n-1}$
$a_n-a_{n-1}=0$

let, $a_n = x$
$x -1 = 0$
$x = 1$

$Homogeneous \ Solution=d*1^n = d$.......................................$(2)$

$ Particular\ Solution :$
Here, $F(x) = 6n^2 + 2n$                                  // Quadratic

let us assume, $a_n = (an^2 + bn + c)*n$         // here root of homogeneous solution is 1 so we have to multiply General quadratic solution by n. ........................$(3)$

$6n^2+ 2n = a_n - a_{n-1}$

$={n*(an^2 + bn + c) - (n-1)*(a(n-1)^2 + b(n-1) + c)}$
$= an^3 + bn^2 + cn - ( an^3 - a -3an^2 +3an + bn^2 + b - 2bn +cn - c )$
$= ( an^3 + bn^2 + cn - an^3 + a +3an^2 -3an - bn^2 - b + 2bn - cn + c )$
$= 3an^2 + (2b-3a)n + (a-b+c)$

Apply Principle of Homogeneity,

$3a =6$                $2b -3a= 2$                    $a-b+c =0$
 $a = 2$                        $b = 4$                                       $c = 2$

Now put values in equation $(3)$,

$Particular\ Solution = n*(2n^2 + 4n +2) = 2n^3 + 4n^2 + 2n$ ..................$(4)$

from equation $(1), (2)$ and $(4)$

Solution of recurrence  $= 2n^3 + 4n^2 + 2n +d$

here $a(1) = 8$                                                                 //given
by putting $n = 1, d=0$

Final Solution of recurrence $= 2n^3 + 4n^2 + 2n = 2n(n+1)^2$

Value of $a(99) = 2×99×(10)^4 = 198*10^4$

So, $K$ value is $198$
edited by
39 votes
39 votes

Here's the easiest way to solve this:

Taking $a_{n-1}$ to LHS,

$a_{n} - a_{n-1} = 6n^{2} + 2n$

Taking summation,

$\sum a_{n} - \sum a_{n-1} = 6 \sum n^{2} + 2 \sum n = (n)(n+1) (2n+1) + n(n+1)$

 

LHS becomes:

 $\sum a_{n} - \sum a_{n-1}= (a_{n}+\sum a_{n-1}) - \sum a_{n-1}=a_{n}$

so our equation reduces to:

$a_{n} = (n)(n+1) (2n+1) + n(n+1)$

Putting n = 99 as asked in the question,

$a_{99} = 99*100*199 + 99*100 = 1980000 = K * 10^{4} $

So K =198. Now wasn't this easy.?

---------------------------------------------------------------------------

p.s.:

$1+2+3+.....+n=\sum{n}=\frac{n(n+1)}{2}$

$1^{2}+2^{2}+3^{2}+.....+n^{2}=\sum{n^{2}}=\frac{n(n+1)(2n+1)}{6}$

$1^{3}+2^{3}+3^{3}+.....+n^{3}=\sum{n^{3}}=\left[\frac{n(n+1)}{2}\right]^{2}=\frac{n^{2}(n+1)^{2}}{4}$

edited by
14 votes
14 votes
why is everyone doing this hard way?

This solution is given,but still

convert recurrence relation to an-(an-1)=6n^2+2n

now all you have to do is apply summation to RHS,

6n^2 converts to n(n+1)(2n-1) (6 gets cancelled) and 2n^2 converts to n(n+1)

so final equation is n(n+1)(2n+1)+n(n+1),putting n =99 answer comes as 1980000,  198*10^4
Answer:

Related questions

57 votes
57 votes
17 answers
1
Sandeep Singh asked Feb 12, 2016
25,583 views
The coefficient of $x^{12}$ in $\left(x^{3}+x^{4}+x^{5}+x^{6}+\dots \right)^{3}$ is ___________.
45 votes
45 votes
4 answers
2
85 votes
85 votes
18 answers
4
Sandeep Singh asked Feb 12, 2016
35,084 views
Let $G$ be a complete undirected graph on $4$ vertices, having $6$ edges with weights being $1, 2, 3, 4, 5,$ and $6$. The maximum possible weight that a minimum weight s...