Question

Someone please help me on both the questions!!!

Answer 1

Step-by-step explanation:

al nth term:

numerator:

$= {n}^{2}$

denominator:

$= n + 1$

so nth term :

$= \frac{ {n}^{2} }{n + 1}$

b). 1; 11; 27; 49

first difference: 10; 16; 22

second difference :6

formula for 1st difference:3a+b

formula for 2nd difference:2a

Therefore:

2a=6

a=3

3a+b=10

sub in a

3(3)+b=10

9+b=10

b=1

sub into quadratic formula

${an}^{2} + bn + c$

$(3) {(1)}^{2} + (1)(1) + c = 1$

$3(1) + 1 + c = 1$

$3 + 1 + c = 1$

$4 + c = 1$

$c = - 3$

So Tn:

$3 {n }^{2} + 1n - 3$

Answer 2

Step-by-step explanation:

a)

it is clearly a sequence of the square numbers over the numbers+1.

1/2 = 1²/(1+1)

4/3 = 2²/(2+1)

9/4 = 3²/(3+1)

16/5 = 4²/(4+1)

so, the nth term would be

n²/(n+1)

b)

an² + bn + c creates the sequence with increasing n (n = 1, 2, 3, 4, ...)

since for sequence the first starting value is often not generated by the general expression, we have to focus on n = 2, n = 3 and n = 4, and with that we get 3 equations with 3 variables (a, b, c) that we can solve :

n = 2

a×2² + b×2 + c = 11 = 4a + 2b + c

n = 3

a×3² + b×3 + c = 27 = 9a + 3b + c

n = 4

a×4² + b×4 + c = 49 = 16a + 4b + c

the first equation gives us

c = 11 - 4a - 2b

when we use this in the second equation, we get

27 = 9a + 3b + 11 - 4a - 2b = 5a + b + 11

b = 16 - 5a

now we use both of that in the third equation :

49 = 16a +4(16 - 5a) + 11 - 4a - 2(16 - 5a) =

= 16a + 64 - 20a + 11 - 4a - 32 + 10a =

= 2a + 43

2a = 6

a = 3

b = 16 - 5a = 16 - 5×3 = 16 - 15 = 1

c = 11 - 4a - 2b = 11 - 4×3 - 2×1 = 11 - 12 - 2 = -3