13. And if it were required to find the sum of n terms 1 1 + 十 + &c. 1.2.3 2.3.4 3.4.5 4.5.6 Let z= 1 Then z-- = 2 1.2 + + -&c. to 'n(n+1)' 1 (n+1)(n+2) 2 (n+1)(n+2) 1.2.3 2.3.4 to n terms, by subtraction. 1 Whence 4-2 (n+1)(n+2) 1.2.3 2.34 3.4.5 terms, by division. 1 1 And consequently+3.4+3.4.5 &c. continued to 1 1 n terms 4 2.(n+1)(n+2) sum required. 14. Required the sum (s) of the series 2 2 2 + &c. 3.4.5 Then ·= x( 1 −x+x2 - x3+x+&c.) 1+x And z=(1+x)x(x-x2+x3-x1+x3 &c. Therefore zx, and x-x2 +x3-x2+x5 &c. = 2 sum required. 1 2 3 4 4 8 16 32 15. Required the sum of the series+++ -+&c. 4 8 16 Let x and s Then -=x+2x2+3x3+4x++5x5 &c. And 2 (1-x) ×(x+2x2+3x3+4x4+5x5 &c.) Whence, by multiplication, x+2x2+3x3+ 4x4 &c. 1-2x+x2 16. It is required to find the sum (s) of the series + (1-x)3 3 =x+4x2+9x+16x1+25x5 &c.. And 2 (1-x)3x(x+4x2+9x3+16x4&c.)=x+x2, as will be found by actual multiplication. a a+d mr 17. Required the sum (s) of the series + m (1-x)a a+(a+d)x+(a+2d)x2+(a+3d)x3+(a+4d)x+ &c. And z=(1-x)X {a +(a+d)x+(a+2d) x2+(a+3d)x3 &c. }=(1-x)a+dx, as will appear by actually multiplying by (1-x)3 Therefore z=(1-x)a+dx; and consequently = {a(r = 1) + d} (r-1)a series required. α + in sum of the infinite EXAMPLES. 1. Required the sum of 100 terms of the series 2, 5, 8, 11, 14, &c. Ans. 15050. 2. Required the sum of 50 terms of the series 1+22 Ans. 42925. +3+4+52 &c. S. It is required to find the sum of the series 1+3x+ 8x2+10x+15x4, continued ad infinitum, &c. when x is less than 1. 1 Ans. (1-x) 4. It is required to find the sum of the series 1+4x +10x2+20x3+35x4, &c. continued ad infinitum, when x is less than 1. 1 Ans. (1—x)a 5. It is required to find the sum of the infinite series 1 + 1.3 + +. 3.5 57 7.9 &c. 5 Ans. 1 or 10' 6. Required the sum of 40 terms of the series (1X2) +(3×4)+(5×6)+(7×8) &c. Ans. 86884 2x-1 7. Required the sum of n terms of the series 8. Required the sum of the infinite series 1 1 1 9. Required the sum of the series +-+ ·+. + 10. It is required to find the sum of the n terms of series 1+8x+27x2+64x3+129x4, &c. continued ad infinitum. 11. Required the sum of n terms of the series -+-+ |