Elements of Algebra

CHAP. X. OF ALGEBRA. 113 Powers. Coefficients. 1st - - - - - - - 1, 1 2d - - - - - - - 1, 2, 1 3d - - - - - - - 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, &c. We see then that the tenth power of $a + b$ will be $a^{10} + 10a^9b + 45a^8b^2 + 120a^7b^3 + 210a^6b^4 + 252a^5b^5 + 210a^4b^6 + 120a^3b^7 + 45a^2b^8 + 10ab^9 + b^{10}$. 346. Now, with regard to the coefficients, it must be observed, that for each power their sum must be equal to the number 2 raised to the same power; for let $a = 1$ and $b = 1$, then each term, without the coefficients, will be 1; consequently, the value of the power will be simply the sum of the coefficients. This sum, in the preceding example, is 1024, and accordingly $(1 + 1)^{10} = 2^{10} = 1024$. It is the same with respect to all other powers; thus, we have for the 1st $1+1=2=2^1$, 2d $1+2+1=4=2^2$, 3d $1+3+3+1=8=2^3$, 4th $1+4+6+4+1=16=2^4$, 5th $1+5+10+10+5+1=32=2^5$, 6th $1+6+15+20+15+6+1=64=2^6$, 7th $1+7+21+35+35+21+7+1=128=2^7$, &c. 347. Another necessary remark, with regard to the coefficients, is, that they increase from the beginning to the middle, and then decrease in the same order. In the even powers, the greatest coefficient is exactly in the middle; but in the odd powers, two coefficients, equal and greater than the others, are found in the middle, belonging to the mean terms. The order of the coefficients likewise deserves particular attention; for it is in this order that we discover the means of determining them for any power whatever, without calculating all the preceding powers. We shall here explain this method, reserving the demonstration however for the next chapter. 348. In order to find the coefficients of any power proposed, the seventh for example, let us write the following fractions one after the other: $\frac{7}{1}, \frac{6}{2}, \frac{5}{3}, \frac{4}{4}, \frac{3}{5}, \frac{2}{6}, \frac{1}{7}$. I