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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 __________.
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10 Answers

8 votes
8 votes
K = 198

Recurrence is solved by

an = (2n+1)(n)(n+1) + n(n+1)

a99 = (199)(99)(100) + (99)(100) = 1980000
6 votes
6 votes
$a_{n} - a_{n-1} = 6n^2 + 2n$

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

$a_{3} - a_{2} = 6*(3)^2 + 2*(3)$

$a_{4} - a_{3} = 6*(4)^2 + 2*(4)$

$.$

$.$

$.$

$.$

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

on adding we get

$a_{n} - a_{1} = 6*[2^2 +3^2 + 4^2 + ...+ n^2] + 2*[2 + 3 + 4 + 5 + ...n]$

$a_{n}  = 6*[2^2 +3^2 + 4^2 + ..+ n^2] + 2*[2 + 3 + 4 + 5 + ...n] +a_{1} $

$a_{n}  = 6*[2^2 +3^2 + 4^2 + ...+ n^2] + 2*[2 + 3 + 4 + 5 + ...n] + 8 $

$a_{n}  = 6*[2^2 +3^2 + 4^2 + ...+ n^2] + 2*[2 + 3 + 4 + 5 + ...n] + 6*1^2 + 2*1 $

$a_{n}  = 6*[1^2 + 2^2 +3^2 + 4^2 + ...+ n^2] + 2*[1 + 2 + 3 + 4 + 5 + ...n]  $

$a_{n}  = 6*( n * ( n+1) * (2n+1) )/6 + (2*n*(n+1))/2$

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

$a_{n}  = n * ( n+1) * (2n+1  + 1)$

$a_{n}  = 2*n * ( n+1)^2$

on putting n = 99, we get

$a_{99}  = 2*99 * ( 99+1)^2$

$a_{n}  = 198 * ( 100)^2$

$a_{n}  = 198 * ( 10)^4$

$K = 198$
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5 votes
5 votes
$a_{n} = 6n^{2} + 2n+a_{n-1}\rightarrow(1)$

$a_{1} = 8,a_{0} = 0$

We can rewrite the equation $(1),$ and get

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

$a_{1}-a_{0} = 6(1)^{2} + 2(1)$

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

$a_{3}-a_{2} = 6(3)^{2} + 2(3)$

$a_{4}-a_{3} = 6(4)^{2} + 2(4)$

$\:\:\:\:\:\:\:\:\Large\vdots$

$a_{98}-a_{97} = 6(98)^{2} + 2(98)$

$a_{99}-a_{98} = 6(99)^{2} + 2(99)$

________________________________________________________

$a_{99}-a_{0} = 6\left[1^{2} + 2^{2} + 3^{2} + \dots + 99^{2} \right] + 2\left[1+2+3+\dots + 99 \right]$

$\implies K \times 10^{4} - 0 = 6\left[\dfrac{99\times 100 \times 199}{6}\right]  + 2\left[\dfrac{99\times 100}{2}\right] $

$\implies K \times 10^{4} =9900 \times 199 +9900$

$\implies K \times 10^{4} =9900 (199 +1)$

$\implies K \times 10^{4} =9900 \times 200$

$\implies K \times 10^{4} =1980000$

$\implies K \times 10^{4} =198\times 10^{4}$

$\therefore K = 198$

So, the correct answer is $198.$
4 votes
4 votes

If the relation is like the one given i.e. an - an-1 = f(n), the we can use the formula:

an = a0  + summation(f(n))[from a1 to an]

so here an = a+ summation(6n2 + 2n) and use the summation formula for n and n2.(here a0 = 0, derived using a1)

Answer:

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