possible ways of taking r of the letters*. Hence the number of the products each of r dimensions will be given by putting a=b=c=...=1 in the continued product. Thus the number required is the coefficient of x in (1 + x + x2+...)", that is in (1-x)". Hence This result can be expressed in the form „H=+r-1Cr COR. The number of terms in the expansion of 290. We shall conclude this chapter by solving the following examples. Ex. 1. Find 14, by the binomial theorem, to six places of decimals. √14=√(16-2)=4(1 8 -- 1 1 1 1 1.3 1 2 8 2.4 82 2.4.6 83 =4{1-0625-001953-0001220·0000095 - 0000010} =3.741657. Ex. 2. Shew that, when x is small, (1 − 3x) ̄3 + (1 − 4x) ̄‡ (1-3x)+(1-4x) 3 =1+2 approximately. Since x is small, its square and higher powers may be rejected; and when all powers of x except the first are neglected, the given expansion becomes equal to * An expression for the sum of the homogeneous products will be found in Art. 296, Ex. 4. Ex. 3. Shew that the integral part of (√3+1)2n+1 is (√3 + 1)2n+1 − (√√3 − 1)2n+1. Since 31 is a proper fraction, (3-1)2n+1 must also be a proper fraction. It therefore follows that if (√3+1)2n+1 − (√√3 − 1)2n+1 be an integer, it must be the integral part of (√3+1)2n+1. Now all the irrational terms disappearing. Since the coefficients of all the different powers of 3 in the last expression are integers, it follows that (√3+1) 2n+1 — (√3 − 1)2n+1 is an integer, and is moreover an even integer. By the following method it can be proved that (√3+1)2n+1 − (√3 − 1)2n+1 is an integer divisible by 2n+1. Represent (√3+ 1)2n+1 − (√√3 − 1)2n+1 by I2n+1· Then I1=2; and it will be found that I=20, and also that Hence (√3+1)2+(√3-1)2=8. - 812n+1= {(√3+1)2n+1 − (√3 − 1)2n+1} { (√3 + 1)2 + (√3 − 1)2} = (√√3 +1)2n+3 − (√3 − 1)2n+3 +4 { (√3 + 1)2n−1 − (√√3 − 1)2n−1}; It follows from the last relation that I2n+3 will be an integer if I2n+1 and I2n- are integers. Now we know that I, and I are integers; hence by induction I2n+1 is always an integer. -1 The relation (A) also shews that Iants will be divisible by 2n+2 provided In+1 is divisible by 2+1 and In-1 by 2". Now we know that I is divisible by 21 and I, by 22; hence I must be divisible by 23; and it will then follow that I, must be divisible by 24; and so on, so that I2n+1 is always divisible by 2n+1. Ex. 4. To shew that, if n be any positive integer, Put for x in the identity proved in Art. 255, Ex. 3; then, after reduction, we have Now expand the expressions on the two sides in powers of In bn 1 y 2) (1 + 2) 1 -1 hence the coefficient of on the right is (-1)*[ca-c1 (a+b)*+...... + ( − 1)”c, (a+rb)2 + ...]. Hence (-1)*c, (a+rb) is zero if k < n, and is equal to (−1)nbn \n if k=n. EXAMPLES XXVIII. 1. Find the sum to infinity of each of the following series: |