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# Can anyone help me with my maths work on the nth term ?

Just need someone to go over this for me ### 7 Answers

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• solving the 1st 4 terms

an = n(n + 1)

a₁ = 1(1 + 1) = 2

a₂ = 2(2 + 1) = 6

a₃ = 3(3 + 1) = 12

a₄ = 4(4 + 1) = 20

The 1st 4 terms are 2,6,12,20

What is the 100th term

a₁₀₀ = 100(100 + 1) = 10,100

Is  53 a term in the sequence: No

Which term is 240

240 = n(n + 1)

240 = n^2 + n

n^2+ n - 240 = 0

(n + 16)(n - 15) = 0

n = - 16, n = 15

The 15th term...

Conclusion

What are the 1st four terms? 2,6,12,20

What is the 100th term? 10,000

Is 53 a term in the sequence? NO.

which term is 240? The 15th term

• T(n)=n(n+1), n>=1=>

T(1)=2; T(2)=6; T(3)=12; T(4)=20; T(100)=10100.

Let n(n+1)=53=>

n^2+n-53=0=>

n=[-1+/-sqr(1+4*53)]/2=>

n=[-1+/-sqr(213)]/2 is not an integer.

Thus, 53 is not in the sequence.

n(n+1)=240=>

n^2+n-240=0=>

n=[-1+/-sqr(1+4*240)]/2=>

n=15 or -16(rejected)

Thus T(15)=240.

•  The nth term of a sequence is n(n + 1)

The first four terms are 2, 6, 12, and 20.

The 100th term is 10,100.

Is 53 a term in the sequence? No

240 is the 15th term.

• You are told that:

f(n) = n(n + 1)

What are the first 4 terms?  Solve for f(1), f(2), f(3), f(4):

f(1) = 1(1 + 1) and f(2) = 2(2 + 1) and f(3) = 3(3 + 1) and f(4) = 4(4 + 1)

f(1) = 1(2) and f(2) = 2(3) and f(3) = 3(4) and f(4) = 4(5)

f(1) = 2 and f(2) = 6 and f(3) = 12 and f(4) = 20

----

What's the 100th term?  Solve for f(100):

f(n) = n(n + 1)

f(100) = 100(100 + 1)

f(100) = 100(101)

f(100) = 10100

----

Is 53 a term?  Set f(n) = 53 and solve for n.  If n isn't an integer then the answer is no:

f(n) = n(n + 1)

53 = n(n + 1)

53 = n² + n

0 = n² + n - 53

n = [ -b ± √(b² - 4ac)] / (2a)

n = [ -1 ± √(1² - 4(1)(-53))] / (2 * 1)

n = [ -1 ± √(1 + 212)] / 2

n = (-1 ± √213) / 2

213 is not a perfect square so the roots of this quadratic are irrational.  Therefore "53" is not a value in the sequence.

----

Which term is 240?  Same steps as above except set f(n) = 240:

f(n) = n(n + 1)

240 = n(n + 1)

240 = n² + n

0 = n² + n - 240

n = [ -b ± √(b² - 4ac)] / (2a)

n = [ -1 ± √(1² - 4(1)(-240))] / (2 * 1)

n = [ -1 ± √(1 + 960)] / 2

n = (-1 ± √961) / 2

n = (-1 ± 31) / 2

n = -32/2 and 30/2

n = -16 and 15

Throw out the negative value since n must be an integer greater than zero, so:

n = 15

240 is the 15th term.

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• What are the first 4 terms ?

Take n(n + 1) and substitute n for 1 2 3 and 4

1(1 + 1) = 1 x 2 = 2

2(2 + 1) = 2 x 3 = 6

3(3 + 1) = 3 x 4 = 12

4(4 + 1) = 4 x 5 = 20

The first four terms are 2 6 12 and 20

The 100th term is 100(100 + 1) .. 100 x 101

Which is 10100

Is 53 a term in this sequence ?

NO .. all the terms are even numbers.

Which term is 240 ?

n(n + 1) = 240

15 x 16 = 240

So it is the 15th term.

• To get the first four terms we substitute for n = 1, 2, 3 and 4

i.e. 1(1 + 1) = 2

2(2 + 1) = 6

3(3 + 1) = 12

4(4 + 1) = 20

The 100th term is when n = 100

so, 100(100 + 1) = 10100

All the terms are even, hence 53 being odd is not a term in the sequence.

If the nth term is 240 then we have n(n + 1) = 240

i.e. n² + n = 240

or, n² + n - 240 = 0

so, (n + 16)(n - 15) = 0

Hence, n = 15...i.e. 15th term

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• First four terms are 2, 6, 12, 20.

100th term is 10100.

All terms are even numbers, so 53 won't appear.

240 is the 15th term because 15 x 16 = 240.

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