Question

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

Answer 1

Answer:

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

Step-by-step 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

Answer 2

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

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

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

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

$S_{17} = \dfrac{17^3+17}2 \implies \boxed{S_{17} = 2465}$