Exercises — Formulae — Σ1, Σn, Σn², Σn³ — proofs

What formula? This is the sum of the first $n$ integers with $n=50$. ${\sum }_{k=1}^{50}k=\frac{n(n+1)}{2}=\frac{50\cdot 51}{2}=\frac{2550}{2}=\boxed{1275}.$ Sanity check (Gauss picture): $50$ columns each summing to $51$ gives $2550$, halved is $1275$. ✔

Use $\sum k^2=\frac{n(n+1)(2n+1)}{6}$ with $n=15$: $\frac{15\cdot 16\cdot 31}{6}=\frac{7440}{6}=\boxed{1240}.$ Here $2n+1=31$. ✔

${\sum }_{k=1}^{10}{k}^3={\left[\frac{10\cdot 11}{2}\right]}^2=55^2=\boxed{3025}.$ The "stunning fact": $\sum k=55$, and $55^2=3025$ — same number. The cube-sum is the square of the triangular number. ✔

Move 1 — linearity. Split into building blocks: ${\sum }_{k=1}^{20}(3k^2-2k+5)=3{\sum }_{k=1}^{20}{k}^2-2{\sum }_{k=1}^{20}k+5{\sum }_{k=1}^{20}1.$ Compute each with $n=20$: $\sum k^2=\frac{20\cdot 21\cdot 41}{6}=2870,\sum k=\frac{20\cdot 21}{2}=210,\sum 1=20.$ Combine: $3(2870)-2(210)+5(20)=8610-420+100=\boxed{8290}.$

Move 2 — shifting. The formula only knows how to start at $k=1$, so cut off the missing head $k=1,2,3,4$: ${\sum }_{k=5}^{12}{k}^2={\sum }_{k=1}^{12}{k}^2-{\sum }_{k=1}^4{k}^2.$ ${\sum }_{k=1}^{12}{k}^2=\frac{12\cdot 13\cdot 25}{6}=650,{\sum }_{k=1}^4{k}^2=\frac{4\cdot 5\cdot 9}{6}=30.$ $650-30=\boxed{620}.$ Why $a-1=4$? We want terms $5$ through $12$ kept, so we subtract everything strictly below $5$, i.e. up to $4$.

General form first. Linearity: ${\sum }_{k=1}^n(2k-1)=2{\sum }_{k=1}^{n}k-{\sum }_{k=1}^{n}1=2\cdot \frac{n(n+1)}{2}-n=n(n+1)-n=n^2.$ So the first $n$ odd numbers sum to $\boxed{n^2}$ — a perfect square. At $n=8$: $1+3+5+\cdots +15=8^2=\boxed{64}.$ Picture: stacking $1,3,5,\dots$ L-shaped shells (gnomons) grows a square one ring at a time — that's why the total is $n^2$.

Expand first so each piece is a standard power sum: $k(k+1)=k^2+k.$ ${\sum }_{k=1}^{n}k(k+1)=\sum k^2+\sum k=\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}.$ Factor $\frac{n(n+1)}{6}$ out of both: $=\frac{n(n+1)}{6}[(2n+1)+3]=\frac{n(n+1)(2n+4)}{6}=\frac{n(n+1)(n+2)}{3}.$ So $\sum _{k=1}^{n}k(k+1)=\boxed{\frac{n(n+1)(n+2)}{3}}.$ Check $n=6$: formula gives $\frac{6\cdot 7\cdot 8}{3}=112$. Direct: $2+6+12+20+30+42=112$. ✔ (These are twice the "tetrahedral numbers" — the 3-D pile-of-oranges count.)

Expand the square: $(2k-1{)}^2=4k^2-4k+1.$ ${\sum }_{k=1}^n(2k-1{)}^2=4\sum k^2-4\sum k+\sum 1.$ $=4\cdot \frac{n(n+1)(2n+1)}{6}-4\cdot \frac{n(n+1)}{2}+n.$ $=\frac{2n(n+1)(2n+1)}{3}-2n(n+1)+n.$ Put over $3$ and simplify (let me expand the first term's bracket): $2n(n+1)(2n+1)=2n(2n^2+3n+1)=4n^3+6n^2+2n$. And $2n(n+1)=2n^2+2n$, so $-2n(n+1)+n=-2n^2-n$. Then $\sum =\frac{4n^3+6n^2+2n}{3}-2n^2-n=\frac{4n^3+6n^2+2n-6n^2-3n}{3}=\frac{4n^3-n}{3}=\frac{n(4n^2-1)}{3}.$ Factor $4n^2-1=(2n-1)(2n+1)$: ${\sum }_{k=1}^n(2k-1{)}^2=\boxed{\frac{n(2n-1)(2n+1)}{3}}.$ Check $n=5$: formula $\frac{5\cdot 9\cdot 11}{3}=165$. Direct: $1+9+25+49+81=165$. ✔

Move 3 — the counter's name is irrelevant; expand the cube. Using $(a+b{)}^3=a^3+3a^{2}b+3a{b}^2+b^3$ with $a=2j,b=3$: $(2j+3{)}^3=8j^3+3(4j^2)(3)+3(2j)(9)+27=8j^3+36j^2+54j+27.$ Sum term by term (rename $j\to k$, same thing): ${\sum }_{k=1}^n(2k+3{)}^3=8\sum k^3+36\sum k^2+54\sum k+27\sum 1.$ $=8{\left[\frac{n(n+1)}{2}\right]}^2+36\cdot \frac{n(n+1)(2n+1)}{6}+54\cdot \frac{n(n+1)}{2}+27n.$ $=2n^2(n+1{)}^2+6n(n+1)(2n+1)+27n(n+1)+27n.$ Check $n=4$ numerically (fastest verification): the terms are $j=1,2,3,4\Rightarrow (5{)}^3+(7{)}^3+(9{)}^3+(11{)}^3=125+343+729+1331=\boxed{2528}.$ Formula at $n=4$: $2\cdot 16\cdot 25+6\cdot 4\cdot 5\cdot 9+27\cdot 4\cdot 5+27\cdot 4=800+1080+540+108=2528$. ✔

Choose the right $f$. We want a difference $f(k+1)-f(k)$ equal to $k(k+1)(k+2)$. A product of three consecutive integers is the "difference" of a product of four consecutive integers, scaled. Try $f(k)=\frac{1}{4}(k-1)k(k+1)(k+2).$ Then $f(k+1)-f(k)=\frac{1}{4}[k(k+1)(k+2)(k+3)-(k-1)k(k+1)(k+2)].$ Factor the common $k(k+1)(k+2)$: $=\frac{1}{4}k(k+1)(k+2)[(k+3)-(k-1)]=\frac{1}{4}k(k+1)(k+2)\cdot 4=k(k+1)(k+2).✓$ Telescope. Summing $f(k+1)-f(k)$ from $1$ to $n$ leaves only ends: ${\sum }_{k=1}^{n}k(k+1)(k+2)=f(n+1)-f(1)=\frac{1}{4}n(n+1)(n+2)(n+3)-\frac{1}{4}\cdot 0\cdot 1\cdot 2\cdot 3.$ The lower end $f(1)=\frac{1}{4}(0)(1)(2)(3)=0$, so ${\sum }_{k=1}^{n}k(k+1)(k+2)=\boxed{\frac{n(n+1)(n+2)(n+3)}{4}}.$ Check $n=5$: formula $\frac{5\cdot 6\cdot 7\cdot 8}{4}=\frac{1680}{4}=420$. Direct: $6+24+60+120+210=420$. ✔ Look at Figure 1 — each term is a "slab," and telescoping is the tower of slabs collapsing so only the outer shell (the four-factor product) survives.

Why partial fractions? We need each term as a difference $f(k)-f(k+1)$ so it telescopes. Split: $\frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1}.$ (Check: $\frac{1}{k}-\frac{1}{k+1}=\frac{(k+1)-k}{k(k+1)}=\frac{1}{k(k+1)}$ ✔.) Telescope with $f(k)=\frac{1}{k}$ (note this is $f(k)-f(k+1)$, so the sum is $f(1)-f(n+1)$): ${\sum }_{k=1}^n\left(\frac{1}{k}-\frac{1}{k+1}\right)=1-\frac{1}{n+1}=\frac{n}{n+1}.$ So $\sum _{k=1}^n\frac{1}{k(k+1)}=\boxed{\frac{n}{n+1}}.$ Check $n=100$: $\frac{100}{101}$. As $n\to \infty$ this $\to 1$ — the infinite series converges to $1$. This is the discrete cousin of an integral converging.

Part A — derive $\sum k^3$. Difference: $(k+1{)}^4-k^4=4k^3+6k^2+4k+1.$ Sum $1\to n$; left telescopes to $(n+1{)}^4-1$: $(n+1{)}^4-1=4\sum k^3+6\sum k^2+4\sum k+n.$ Substitute $\sum k^2=\frac{n(n+1)(2n+1)}{6}$ and $\sum k=\frac{n(n+1)}{2}$: $4\sum k^3=(n+1{)}^4-1-n(n+1)(2n+1)-2n(n+1)-n.$ Expand $(n+1{)}^4-1=n^4+4n^3+6n^2+4n$; $n(n+1)(2n+1)=2n^3+3n^2+n$; $2n(n+1)=2n^2+2n$. Then $4\sum k^3=n^4+4n^3+6n^2+4n-(2n^3+3n^2+n)-(2n^2+2n)-n=n^4+2n^3+n^2=n^2(n+1{)}^2.$ $\therefore {\sum }_{k=1}^n{k}^3=\frac{n^2(n+1{)}^2}{4}={\left[\frac{n(n+1)}{2}\right]}^2.✓$ Part B — odd cubes. Trick: $\text{(all cubes)}-\text{(even cubes)}$ up to $2m-1$. Even cubes are ${\sum }_{j=1}^{m-1}(2j{)}^3=8{\sum }_{j=1}^{m-1}{j}^3$. For $m=4$ we want $1^3+3^3+5^3+7^3$: all cubes to $7$ minus even cubes $2^3,4^3,6^3$. ${\sum }_{k=1}^7{k}^3={\left[\frac{7\cdot 8}{2}\right]}^2=28^2=784.$ $2^3+4^3+6^3=8(1^3+2^3+3^3)=8\cdot 36=288.$ $1^3+3^3+5^3+7^3=784-288=\boxed{496}.$ Direct check: $1+27+125+343=496$. ✔

Expand the product to reach standard sums: $(3k-1)(3k+2)=9k^2+6k-3k-2=9k^2+3k-2.$ Linearity: ${\sum }_{k=1}^n(9k^2+3k-2)=9\sum k^2+3\sum k-2\sum 1.$ $=9\cdot \frac{n(n+1)(2n+1)}{6}+3\cdot \frac{n(n+1)}{2}-2n=\frac{3n(n+1)(2n+1)}{2}+\frac{3n(n+1)}{2}-2n.$ Combine the first two over $2$: $\frac{3n(n+1)[(2n+1)+1]}{2}=\frac{3n(n+1)(2n+2)}{2}=3n(n+1{)}^2$. So ${\sum }_{k=1}^n(3k-1)(3k+2)=3n(n+1{)}^2-2n.$ At $n=10$: $3\cdot 10\cdot 121-20=3630-20=\boxed{3610}.$ Spot-check the general form at $n=1$: formula $3\cdot 1\cdot 4-2=10$; direct $(3-1)(3+2)=2\cdot 5=10$ ✔.

Exact sum over the model. $\frac{\sum k^2}{n^3/3}=\frac{3}{n^3}\cdot \frac{n(n+1)(2n+1)}{6}=\frac{(n+1)(2n+1)}{2n^2}=\frac{2n^2+3n+1}{2n^2}=1+\frac{3}{2n}+\frac{1}{2n^2}.$ At $n=10$: $1+0.15+0.005=\boxed{1.155}.$ At $n=1000$: $1+0.0015+0.0000005=\boxed{1.0015005}.$ Limit: as $n\to \infty$ both correction terms vanish, so the ratio $\to \boxed{1}$. That confirms $\sum k^2\sim \frac{1}{3}{n}^3$. Why the integral? Scaling $k=nx$, $\frac{1}{n^3}\sum k^2\approx \frac{1}{n}\sum {\left(\frac{k}{n}\right)}^2$ is a Riemann sum for ${\int }_0^1{x}^{2}dx=\frac{1}{3}$. The discrete power-sum is a staircase approximation to the smooth area — the $\frac{3}{2n}$ term is the leftover "staircase overshoot" that melts away. See Definite Integrals as Limits of Sums.

L1: grab the right formula, plug, compute. L2: linearity + shifting. L3: expand/re-index before any formula bites. L4: invent the telescoping $f(k)$ yourself. L5: pick the whole strategy — split odd/even, compare to integrals, and know the difference between "$\sim$" and "$=$".