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Consider the recurrence relation a1 = 8, an = 6n2 + 2n + an-1. Let a99 = K x 104 . The value of K is __________.
- a)198 × 104
- b)194 × 104
- c)192 × 104
- d)196 × 104
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
Consider the recurrence relation a1 = 8, an = 6n2 + 2n + an-1.Let a99 ...
an = 6n2 + 2n + an−1
= 6n2 + 2n + 6(n − 1)2 + 2(n − 1) + an−2
= 6n2 + 2n + 6(n − 1)2 + 2(n − 1) + 6(n − 2)2 + 2(n − 2)+. . . . . . +a1
= 6n2 + 2n + 6(n − 1)2 + 2(n − 1) + 6(n − 2)2 + 2(n − 2)+. . . . . . +6.12 + 2.1
= 6(n2 + (n − 1)2 +. . . +22 + 12) + 2(n + (n − 1)+. . . +2 + 1)
for n = 99 a99 = 2 x 99 x (99+1)2 = 198 x 104
= 6n2 + 2n + 6(n − 1)2 + 2(n − 1) + an−2
= 6n2 + 2n + 6(n − 1)2 + 2(n − 1) + 6(n − 2)2 + 2(n − 2)+. . . . . . +a1
= 6n2 + 2n + 6(n − 1)2 + 2(n − 1) + 6(n − 2)2 + 2(n − 2)+. . . . . . +6.12 + 2.1
= 6(n2 + (n − 1)2 +. . . +22 + 12) + 2(n + (n − 1)+. . . +2 + 1)
for n = 99 a99 = 2 x 99 x (99+1)2 = 198 x 104Most Upvoted Answer
Consider the recurrence relation a1 = 8, an = 6n2 + 2n + an-1.Let a99 ...
an = 6n2 + 2n + an−1
= 6n2 + 2n + 6(n − 1)2 + 2(n − 1) + an−2
= 6n2 + 2n + 6(n − 1)2 + 2(n − 1) + 6(n − 2)2 + 2(n − 2)+. . . . . . +a1
= 6n2 + 2n + 6(n − 1)2 + 2(n − 1) + 6(n − 2)2 + 2(n − 2)+. . . . . . +6.12 + 2.1
= 6(n2 + (n − 1)2 +. . . +22 + 12) + 2(n + (n − 1)+. . . +2 + 1)
for n = 99 a99 = 2 x 99 x (99+1)2 = 198 x 104
= 6n2 + 2n + 6(n − 1)2 + 2(n − 1) + an−2
= 6n2 + 2n + 6(n − 1)2 + 2(n − 1) + 6(n − 2)2 + 2(n − 2)+. . . . . . +a1
= 6n2 + 2n + 6(n − 1)2 + 2(n − 1) + 6(n − 2)2 + 2(n − 2)+. . . . . . +6.12 + 2.1
= 6(n2 + (n − 1)2 +. . . +22 + 12) + 2(n + (n − 1)+. . . +2 + 1)
for n = 99 a99 = 2 x 99 x (99+1)2 = 198 x 104Free Test
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Community Answer
Consider the recurrence relation a1 = 8, an = 6n2 + 2n + an-1.Let a99 ...
To find the value of K, we need to find the 99th term of the sequence.
Using the given recurrence relation, we have:
a1 = 8
a2 = 6(2^2) - 2(2) + 8 = 20
a3 = 6(3^2) - 2(3) + 20 = 48
a4 = 6(4^2) - 2(4) + 48 = 88
a5 = 6(5^2) - 2(5) + 88 = 140
...
We can observe that the nth term can be written as:
an = 2n^3 - 2n + 6(n-1) + 2
By substituting n = 99 into this equation, we get:
a99 = 2(99^3) - 2(99) + 6(98) + 2
Calculating this expression, we find:
a99 = 2(970299) - 198 + 588 + 2
= 1940598
Therefore, K = 1940598 / 10^4 = 194.0598 ≈ 194.
So the value of K is 194.
Using the given recurrence relation, we have:
a1 = 8
a2 = 6(2^2) - 2(2) + 8 = 20
a3 = 6(3^2) - 2(3) + 20 = 48
a4 = 6(4^2) - 2(4) + 48 = 88
a5 = 6(5^2) - 2(5) + 88 = 140
...
We can observe that the nth term can be written as:
an = 2n^3 - 2n + 6(n-1) + 2
By substituting n = 99 into this equation, we get:
a99 = 2(99^3) - 2(99) + 6(98) + 2
Calculating this expression, we find:
a99 = 2(970299) - 198 + 588 + 2
= 1940598
Therefore, K = 1940598 / 10^4 = 194.0598 ≈ 194.
So the value of K is 194.
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Consider the recurrence relation a1 = 8, an = 6n2 + 2n + an-1.Let a99 = K x 104 . The value of K is __________.a)198 × 104b)194× 104c)192× 104d)196× 104Correct answer is option 'A'. Can you explain this answer?
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