Question

Find the value of a4 for the recurrence relation an=2an-1+3, with a0=6. 65 141 221 320

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

GIVEN

A recurrence relation is given by

$\sf{a_n = 2 a_{n-1} + 3 \: \: \: \: \: \: \: with \: \: \: a_0 = 6}$

TO CHOOSE THE CORRECT OPTION

$\sf{a_4 = }$

• 65

• 141

• 221

• 320

CALCULATION

Here the recurrence relation is given by

$\sf{a_n = 2 a_{n-1} + 3 \: \: \: \: \: \: \: with \: \: \: a_0 = 6}$

So

$\sf{a_0 = 6}$

$\sf{a_1 = 2 a_{0} + 3 \: = (2 \times 6) + 3 = 15 }$

$\sf{a_2 = 2 a_{1} + 3 \: = (2 \times 15) + 3 = 33 }$

$\sf{a_3 = 2 a_{2} + 3 \: = (2 \times 33) + 3 = 69 }$

$\sf{a_4 = 2 a_{3} + 3 \: = (2 \times 69) + 3 = 141 }$

RESULT

$\boxed{ \sf{\: \: \: Hence \: \: \: \: a_4 = 141 \: \: \: }}$

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LEARN MORE FROM BRAINLY

Solve the recurrence relation

s(k)-10s(k-1)+9s(k-2)=0 with s(0)=3,s(1)=11

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Answer 2

Answer:

141

Step-by-step explanation:

a

n

=2a

n−1

+3witha

0

=6

So

\sf{a_0 = 6}a

0

=6

\sf{a_1 = 2 a_{0} + 3 \: = (2 \times 6) + 3 = 15 }a

1

=2a

0

+3=(2×6)+3=15

\sf{a_2 = 2 a_{1} + 3 \: = (2 \times 15) + 3 = 33 }a

2

=2a

1

+3=(2×15)+3=33

\sf{a_3 = 2 a_{2} + 3 \: = (2 \times 33) + 3 = 69 }a

3

=2a

2

+3=(2×33)+3=69

\sf{a_4 = 2 a_{3} + 3 \: = (2 \times 69) + 3 = 141 }a

4

=2a

3

+3=(2×69)+3=141