Paduaruschelleidn
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find the next four terms in the sequence 1,4,9,16,25,__,___,__,__,​

Sagot :

Dennispark872idn

Answer:

36,49,66,85

Step-by-step explanation:

just add 2

sana makatulong

AGboiiidn

[tex] \large\underline \mathcal{{QUESTION:}}[/tex]

  • find the next four terms in the sequence 1,4,9,16,25,__,___,__,__,

[tex] \large\underline \mathcal{{ANSWER:}}[/tex]

  • The next four terms are 36,49,64,and 81

[tex]\\[/tex]

[tex] \large\underline \mathcal{{SOLUTION:}}[/tex]

First , We will solve for the nth term or the formula of the quadratic sequence.

  • an²+bn+c

[tex]\\[/tex]

Using the formula , the first term is 1 , so we will substitute the value of 1 to n. Then equate to the first term or (1)

[tex] \sf an²+bn+c=1 \\ \sf a(1)²+b(1)+c=1 \\ \boxed{a+b+c=1} \: \longleftarrow \mathtt{eq1}[/tex]

Again , Substitute 2 to n and equate to the second term or (4)

[tex] \sf an²+bn+c=4 \\ \sf a(2)²+b(2)+c=4 \\ \boxed{4a+2b+c=4} \: \longleftarrow \mathtt{eq2}[/tex]

Lastly , Substitute 3 to n and equate to the third term or (9)

[tex] \sf an²+bn+c=9 \\ \sf a(3)²+b(3)+c=9 \\\boxed{9a+3b+c=9} \: \longleftarrow \mathtt{eq3}[/tex]

[tex]\\[/tex]

Now that we got the three equations we will subtract them.

Equation 2 - Equation 1

[tex] \: \: \: \: \: \: \: 4a+2b+c=4\\-\underline{\: \: \: \: \: \: \: \: a+b+c=1} \\ = \boxed{3a + b = 3}[/tex]

Equation 3 - Equation 2

[tex] \: \: \: \: \: 9a+3b+c=9 \\-\underline{ \: \: 4a+2b+c=4} \\ = \boxed{5a + b = 5}[/tex]

[tex]\\[/tex]

Solving the value of a to the differences of the equation by elimination:

[tex] \: \: \: \: \: 5a+b=5 \\\underline{- \: \: \: 3a+b=3} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: 2a = 2 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \:\frac{2a}{2} = \frac{2}{2} \\ \\ \: \implies \boxed{a = 1}[/tex]

[tex]\\[/tex]

Now , we get the value of a , thus we will find for the values of b and c. Let's substitute the value of a to the equation:

[tex] \: \: \: \: \: 5(1)+b=5 \\\underline{ + \: \: \: 3(1)+b=3} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: 8 +2 b = 8 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 2b = 8 - 8 \\ \: \: \: \: \: \: \: 2b = 0 \\ \\ \: \implies\boxed{b = 0}[/tex]

[tex]\\[/tex]

Lastly , We will substitute to any of the first equation:

[tex] \: \: \: \: \: \: \: \: \:a + b + c = 1 \\ (1) + (0) + c = 1 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 1 + c = 1 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: c = 1 - 1 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{c = 0}[/tex]

[tex]\\[/tex]

a = 1 , b = 0 , c = 0

Summing up , We get the formula:

  • an²+bn+c
  • 1n²+0(n)+0
  • n²

[tex]\\[/tex]

Now that we got the formula: We will find for the next four terms:

[tex]\\[/tex]

Substitute n = 6

  • n² = 6² = 36

[tex]\\[/tex]

Substitute n = 7

  • n² = 7² = 49

[tex]\\[/tex]

Substitute n = 8

  • n² = 8² = 64

[tex]\\[/tex]

Substitute n = 9

  • n² = 9² = 81

[tex]\\[/tex]

  • Therefore , The sequence is 1,4,9,16,25,36,49,64,81
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