Sum a block of odd squares

It is given that ∑r=1nr2=n(n+1)(2n+1)6\sum_{r=1}^nr^2=\frac{n(n+1)(2n+1)}6.

(i) Show ∑r=1n(2r−1)2=n(4n2−1)3\sum_{r=1}^n(2r-1)^2=\frac{n(4n^2-1)}3. marks

(ii) Hence find an expression for 132+152+⋯+(4m−1)213^2+15^2+\cdots+(4m-1)^2 in terms of mm. marks

[2024/MI/Prelim/I/9b]

Total marks: 7

Solution

(i)

Expand and sum

Given ∑r=1nr2=n(n+1)(2n+1)6\sum_{r=1}^nr^2=\frac{n(n+1)(2n+1)}6, show that ∑r=1n(2r−1)2=n(4n2−1)3\sum_{r=1}^n(2r-1)^2=\frac{n(4n^2-1)}3.

Explore this idea

Hint

  1. Expand (2r−1)2(2r-1)^2.

  2. Split into ∑r2\sum r^2, ∑r\sum r and ∑1\sum 1 and substitute the standard results.

Answer · (i)

n(4n2−1)/3n(4n^2-1)/3.

Explore the idea

Odd squares, one bar each

The bars are 1, 9, 25, 49,…1,\ 9,\ 25,\ 49,\dots. Drag the upper limit and compare the direct total with the formula.

Upper limitr = 1 to 5
Try an example
Σ (2r−1)², r = 1 to 5
165
Number of terms
5
n(4n²−1)/3
165
matches the direct total
Key idea

The sum of the first nn odd squares is a cubic in nn, because the terms themselves are quadratic.

Ideas to try
  1. At n=3n=3 the total is 1+9+25=351+9+25=35 and 3(36−1)3=35\frac{3(36-1)}3=35.
  2. Switch to the running total: it is the cubic-looking curve n(4n2−1)3\frac{n(4n^2-1)}3.
(ii)

Subtract partial sums

Hence find an expression, in terms of mm, for 132+152+⋯+(4m−1)213^2+15^2+\cdots+(4m-1)^2.

Explore this idea

Hint

  1. Match 1313 and 4m−14m-1 with 2r−12r-1 to find the limits.

  2. Write the block as a difference of two sums that both start at 11.

Answer · (ii)

2m(16m2−1)/3−2862m(16m^2-1)/3-286.

m∈Zm\in\mathbb Z, m≥4m\ge4.

Explore the idea

A block is a difference of two towers

Move both limits. The closed form is the tower up to the upper limit minus the tower below the lower limit.

Limits of the sumr = 7 to 10
Try an example
Σ (2r−1)², r = 7 to 10
1044
Number of terms
4
S(upper) − S(lower − 1)
1044
matches the direct total
Key idea

Any block ∑r=ab\sum_{r=a}^{b} equals Sb−Sa−1S_b-S_{a-1}. Forgetting the −1-1 would drop the first term of the block.

Ideas to try
  1. Set the lower limit to 77. The grey bars are the 66 terms subtracted: 12+⋯+112=2861^2+\cdots+11^2=286.
  2. Slide the lower limit left and watch the formula change by exactly the term you add.
Takeaway

Two sums make a block

Closed forms start at r=1r=1. To sum from r=ar=a to r=br=b, subtract the sum to a−1a-1 from the sum to bb.