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Let S1= 1, S2=2+3, S3= 4+5+6
Find S7
Find S17
Find Sn

User Jprice
by
2.8k points

2 Answers

8 votes
8 votes

Each term Sₙ consists of the sum of a triangular number of terms, which are given by


T_n = \displaystyle \sum_(k=1)^n k = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}2

The triangular numbers are given recursively for n ≥ 1 by


T_n = T_(n-1) + n

starting with T₀ = 0.

For example,

• S₁ = 1 and


\displaystyle S_1 = \sum_(k=T_0+1)^(T_1) k = \sum_(k=1)^1 k = 1

• S₂ = 2 + 3 and


\displaystyle S_2 = \sum_(k=T_1+1)^(T_2) k = \sum_(k=2)^3 k = 2 + 3

• S₃ = 4 + 5 + 6 and


\displaystyle S_3 = \sum_(k=T_2+1)^(T_3) k = \sum_(k=4)^6 k = 4 + 5 + 6

Then the n-th term of the sequence we're considering is


S_n = \displaystyle \sum_{k=T_(n-1)+1}^(T_n) k = \sum_{k=T_(n-1)+1}^{T_(n-1)+n} k

Expanding this sum, we have


S_n = \left(T_(n-1)+1\right) + \left(T_(n-1)+2\right) + \left(T_(n-1)+3\right) + \cdots + \left(T_(n-1)+n\right)

There are n terms on the right side, and hence n copies of
T_(n-1), and the rest of the terms make up the next triangular number
T_n :


S_n = nT_(n-1) + 1 + 2 + 3 + \cdots + n


S_n = nT_(n-1) + \displaystyle \sum_(k=1)^n k


S_n = nT_(n-1) + T_n

We have a closed form for
T_n, so we end up with


S_n = n \cdot \frac{(n-1)n}2 + \frac{n(n+1)}2 \implies \boxed{S_n=\frac{n^3+n}2}

From here it's easy to find S₇ and S₁₇.


S_7 = \frac{7^3+7}2 \implies \boxed{S_7 = 175}


S_(17) = \frac{17^3+17}2 \implies \boxed{S_(17) = 2465}

User Tspore
by
3.1k points
21 votes
21 votes

Answer:

  • S7 = 175
  • S17 = 2465
  • Sn = 1/2(n³ +n)

Explanation:

The progression of sums is ...

1, 5, 15, 34, 65, ...

So, first differences are ...

4, 10, 19, 31

Second differences are ...

6, 9, 12, ...

Third differences are constant:

3, 3, ...

This means the expression for Sn will be a cubic expression. If dn is the first of the n-th differences, then the equation can be written as ...

Sn = S1 +(n -1)(d1 +(n -2)/2(d2 +(n -3)/3(d3)))

And this simplifies a little bit to ...

Sn = 1 +(n -1)(4 +(n -2)(n +3)/2)

In simpler form, we have ...

Sn = 1/2(n³ +n)

Then the two terms we're interested in are ...

S7 = (1/2)(7³ +7) = 175

S17 = (1/2)(17³ +17) = 2465

User Scarlet
by
3.5k points