ch1.2.10-higher-powers-compound

Ch 1.2.10 — Of the Higher Powers of Compound Quantities

Summary: Extends the study of binomial powers beyond squares and cubes to arbitrary degree, builds the Pascal’s triangle coefficient table, and derives a direct formula for any coefficient without computing all preceding powers.

Sources: chapter-1.2.10

Last updated: 2026-04-29

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Overview (articles 340–341)

After treating squares in ch1.2.6-squares-of-compound-quantities and cubes in ch1.2.9-cubes-and-cube-root-extraction, Euler generalises to any integer power. For a compound root, the exponent is written outside parentheses: $(a+b{)}^5$, $(a-b{)}^6$, etc.

He then computes $(a+b{)}^n$ by repeated multiplication up to $n=6$:

$$(a+b{)}^2=a^2+2ab+b^2$$ $$(a+b{)}^3=a^3+3a^{2}b+3a{b}^2+b^3$$ $$(a+b{)}^4=a^4+4a^{3}b+6a^2{b}^2+4a{b}^3+b^4$$ $$(a+b{)}^5=a^5+5a^{4}b+10a^3{b}^2+10a^2{b}^3+5a{b}^4+b^5$$ $$(a+b{)}^6=a^6+6a^{5}b+15a^4{b}^2+20a^3{b}^3+15a^2{b}^4+6a{b}^5+b^6$$

Powers of $a-b$ (article 342)

The expansion of $(a-b{)}^n$ has the same absolute terms as $(a+b{)}^n$, with the sign rule: even-position terms keep $+$, odd-position terms (2nd, 4th, …) get $-$. The reason: $(-b{)}^k$ alternates sign, giving $-b,+b^2,-b^3,+b^4,\dots$ (source: chapter-1.2.10)

Structure of Terms (article 344)

Ignoring coefficients, every term in $(a+b{)}^n$ has the form $a^{n-k}{b}^k$ for $k=0,1,\dots ,n$. The sum of exponents is always $n$. The terms march from $a^n$ down to $b^n$ as $k$ increases. (source: chapter-1.2.10)

Pascal’s Triangle (article 345)

The coefficients form a triangular table:

Power	Coefficients
1st	1, 1
2nd	1, 2, 1
3rd	1, 3, 3, 1
4th	1, 4, 6, 4, 1
5th	1, 5, 10, 10, 5, 1
6th	1, 6, 15, 20, 15, 6, 1
7th	1, 7, 21, 35, 35, 21, 7, 1
8th	1, 8, 28, 56, 70, 56, 28, 8, 1
9th	1, 9, 36, 84, 126, 126, 84, 36, 9, 1
10th	1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1

The first and last coefficient in every row is 1. The second coefficient equals the exponent. Coefficients are symmetric (palindromic rows). See pascal-triangle. (source: chapter-1.2.10)

Sum of Coefficients = $2^n$ (article 346)

Setting $a=b=1$ makes every term equal 1, so $(1+1{)}^n=2^n$ equals the sum of all coefficients in row $n$. Euler verifies this for rows 1 through 7. (source: chapter-1.2.10)

Symmetry of Coefficients (article 347)

Coefficients increase from the first term to the middle, then decrease in the same order. For even powers there is a single largest term in the exact middle; for odd powers, two equal largest terms flank the middle. (source: chapter-1.2.10)

Direct Coefficient Formula (articles 348–351)

To find all coefficients of $(a+b{)}^n$ directly, write the sequence of fractions

$$\frac{n}{1},\frac{n-1}{2},\frac{n-2}{3},\dots ,\frac{1}{n}$$

Then:

• 1st coefficient = $1$
• 2nd coefficient = $\frac{n}{1}$
• $k$th coefficient = product of the first $k-1$ fractions

For the 7th power the fractions are $\frac{7}{1},\frac{6}{2},\frac{5}{3},\frac{4}{4},\frac{3}{5},\frac{2}{6},\frac{1}{7}$, giving coefficients $1,7,21,35,35,21,7,1$.

In modern notation the $k$th coefficient of $(a+b{)}^n$ is the binomial coefficient

$$\left(\genfrac{}{}{0ex}{}{n}{k-1}\right)=\frac{n(n-1)\cdots (n-k+2)}{1\cdot 2\cdots (k-1)}$$

Euler demonstrates the formula applied to the 10th power, and notes it extends to any exponent, exhibiting $(a+b{)}^{100}$ explicitly. See binomial-theorem. (source: chapter-1.2.10)

Related pages

• pascal-triangle
• binomial-theorem
• ch1.2.6-squares-of-compound-quantities
• ch1.2.9-cubes-and-cube-root-extraction
• ch1.2.11-transpositions-and-binomial-proof
• ch1.2.12-irrational-powers-infinite-series
• algebraic-identities
• powers-and-exponents