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mina [271]
2 years ago
5

b) The first five terms in a different sequence are -29, -26, -21, -14, -5. Find, in terms of n, a formula for the nth term, V,

of the sequence. ​
Mathematics
1 answer:
oksano4ka [1.4K]2 years ago
3 0

If v_n is the n-th term in the sequence, observe that

v_2 - v_1 = -26 - (-29) = 3

v_3 - v_2 = -21 - (-26) = 5

v_4 - v_3 = -14 - (-21) = 7

v_5 - v_4 = -5 - (-14) = 9

and if the pattern continues,

v_n - v_{n-1} = 2n - 1

so the sequence is defined recursively by

\begin{cases} v_1 = -29 \\ v_n = v_{n-1} + 2n - 1 & \text{for } n > 1 \end{cases}

By this definition,

v_{n-1} = v_{n-2} + 2(n-1) - 1 = v_{n-2} + 2n - 3

v_{n-2} = v_{n-3} + 2(n-2) - 1 = v_{n-3} + 2n - 5

and so on. Then by substitution, we have

v_n = v_{n-1} + 2n - 1

v_n = (v_{n-2} + 2n - 3) + 2n - 1 = v_{n-2} + 2\times2n - (1 + 3)

v_n = (v_{n-3} + 2n-5) + 2\times2n - (1 + 3) = v_{n-3} + 3\times2n - (1 + 3 + 5)

and if we keep doing this we'll eventually get v_n in terms of v_1 to be

v_n = v_1 + (n-1)\times2n - (1 + 3 + 5 + \cdots + (2(n-1)-1))

Evaluate the sum:

Let

S = 1 + 3 + 5 + \cdots + (2(n-1)-1) = 1 + 3 + 5 + \cdots + (2n-3)

S' = 2 + 4 + 6 + \cdots + (2n - 2)

Then

S + S' = 1 + 2 + 3 + 4 + \cdots + (2n-3) + (2n-2) = \displaystyle \sum_{k=1}^{2n-2} k

Recall that

\displaystyle \sum_{i=1}^n i = 1 + 2 + 3 + \cdots + n = \dfrac{n(n+1)}2

so that

S + S' = \dfrac{(2n-2)(2n-1)}2

and

S' = 2 + 4 + 6 + \cdots + (2n-2) = 2 \left(1 + 2 + 3+ \cdots + (n-1)\right) = (n-1)n

So, we find

S = (S + S') - S' = \dfrac{(2n-2)(2n-1)}2 - n(n-1) = (n-1)^2

Then the n-th term to the sequence is

v_n = v_1 +2n(n-1) - S = -29 + 2n^2 - 2n - (n-1)^2 = \boxed{n^2-30}

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