125. Required the sum (s) of the circulating decimal 999999... continued ad infinitum. Or, 10 100 +1000+ 10000 +&c.: + 126. Required the sum (s) of the series a2+(a+d)2+(a+2d)2+ (a+3d)2+(a +4d)2+ &c. continued to n terms. Here aa; (a+d)2=a2+2× lad+ld2 (a+2d)2=a2+2×2ad+4d; (a+3d)2=a2+2×3ad+9ď; and (a+4d)2=a2+2×4ad+16ď2; whence 8= Sum of n terms of (1+1+1+1+ &c.)a2 キ . . . ditto of (0+1+2+3+ &c.)2ad But n terms of 1+1+1+1+&c. = n. And of 0+1+2+3+4 n(n-1) 1.2 n(n-1)(2n-1) 1.2.3 Also of 0+1+4+ 9 + 16 + 25+ &c '; ... s=na2+n(n-1)ad+ the whole sum of the series to n terms. 127. Required the sum (s) of the series a3+(a+d)3+(a+2d)3 +(a+3d)+(a+4d)+ &c. continued to n terms. Here a3 a (a+ d)3=a3+3×1a2d+3× laď2+ 1ď (a+2d)=a3x2a'd+3X 4ad2 + 8ď3 (a+3d)3=a3+3×3a2d+3× 9ad2+ 27ď3 Whence s (a+4d)3 a3+3x4a2d+3X16ad264d3 But n terms of 1+1+1+1— &c. = n. Ditto ..... n(n-1)(2n-1) ..... of 0+1+4+9+ &c.= 2.3 Ditto.... of 0+1+8+27+ &c. = (n—1)1+2(n—1)3+(n—1)3 4 128. Required the sum (s) of n terms of the series 1+3+7+ 15-31 &c. Here the terms of the series are, evidently, equal to 1, (1+2), (1+2+4), (1+2+4+8), &c. or to the successive sums of the geometrical series, 1, 2, 4, 8, 16, &c. Let, therefore, a=1 and r=2, and we shall have atar+ar2+ar3+ar1+ &c.=1+2+4+8+ 16 + &c. But the successive sums of 1, 2, 3, 4, &c. terms of this series are a 2. r-1 a a a -1. But a+rofr3+r*+&c.=(r”—1) ×———, and 1+1+1+1+1 X -1' 3 17 15 31 129. Sum n terms of the series+++ + -&c. 8 16 Here, the terms of the series are the successive sums of the ge The successive sums of 1, 2, 3, 4, &c. terms of this series are, (n-1)2"+1 = 27-1 =sum required. 130. Required the sum of the infinite series of the reciprocals. of the triangular numbers +++i+is+&c. Let +is+is+&c. ad infinitum = s. 1 1 1 1 Or +1.3 +23 +2.5 +3.5 + &c. .... = s. 1.1 1 1 Then + &c.. .... + 5 +&c. 3 1 3 1 + + + + &c. 5 6 1 1 1 ; or s2 sum required. 131. Find the sum of ʼn terms of the same series, + § ÷ 1 + +15+ &c. Let s=1+2 to t2 to t&c. to-. 1 1 1 1 4 5 2 4 5 1 10 1 1 Then s +++&c. to . . to And s + n Therefore, by subtracting 1 1 1 6. 12 20 1 1 1 1 n toot &c. 2 Then s = 3 + 3 + 4 ++&c. by transposition. 1 And 1: 1.2+2.3 +3.4 +45 + &c. by subtraction. sum of n terms of the 1 1 1 +&c. 2 1.2.3 +2.3.4 +3.4.5 +4.5.6 + &c. But÷2=4 1 1 1 1.2.3 +2.3.4 +3.4.5 by dividing each side by 2. 1 1 1 123 +2.3.4 +3.4.5 +4.5.6 +5.6.7 4, which is the sum required. &c. 133. And if it were required to find the sum of n terms of the 1 1 1 terms. (n+1)(x+2) 4.5 +5.6 +6.7 + Therefore 1 7.8 2 ̄(n+1)(n+2)=1.2.3 2.3.4 +3.4.5+ &c. to n terms, by + &c. to n terms, by division. And, consequently, 1.2.3 1 + + &c. continued to ʼn terms = 2.3.4 3.4.5 sum required. 1 1 4 2(n+1)(n+2) 134. Required the sum (s) of the series-+-+6+ &c. continued ad infinitum. Let x= ; in which case s=x-x x3-x2+ &c. Then, if this be multiplied by 1+x, we shall have s(1+x)=(1+x)× (x—x2+x3-x2+x-&c.), where, by performing the operation on the right hand member, there will arise s(1+x)=x. Therefore, sx-x2+x3-x2+x3-&c. 135. Required the sum (s) of the series ++&++ +&c. continued ad infinitum. Let x; in which case s=x+2x2 + 3x3 + 4x2 + &c. Then, if this be multiplied by (1-x), we shall have s(1—x)2=—(1—x)2×(x+2x2+ 3x2+4x2 + &c.) Where, by performing the operation at length, on the right hand member, as before, there will arise s(1-x)2=x Therefore, also, x+2x2+3x3 +4x2+5x+ &c. x (1-x)2' Or 1 2 3 4 5 2 4 8 16 + 32+ &c. 136. Required the sum (s) of the series = sum. (1—1) 2 ++++ 25 &c. continued ad infinitum. ́ Let x = ; in which case s=x+4x2+92 +16x+ &c. Then, if this be multiplied by (1-x)3, we shall have s(1 — - x)3 —(1—x)3× (x+4x2+9x2+16x*+&c.)=x+x2, as will be found by performing the operation. Therefore s(1-x)=x+x2, or s= (1—x)3 (1 —x)2° 3 9 27 81 137. Required the sum (s) of the infinite series +a+da+2da+3d a+4d Let x; in which case s= + -x+ -x2 + &c. Then, if this be multiplied by (1-x)2, we shall have s(1 — x)2. Whence, restoring the value of x, we shall have, by substitu sum of the series, as required. 138. Required the sum of n terms of the series (r−1)2 m(m+r) +(m+r)(m+2r) +(m+2r)(m+3r) +(m+3r)(m+4r) m {a(r=-1) a(r-1)+d |