PROB. 7. To find the sum of n terms of the series a2+a+al2 +a+2d2+a+3a)2+, &c.

First, by actually squaring the terms, we have

Whence 1+1+1+1+&c. (to n terms) x a2
+0+1+2+3+&c. (to n terms) × 2 ad
+0+1+4+9+&c. (to n terms) × d2

n

1+1+1+1+&c.'

But the sum of

0+1+2+3+ &c.

to n terms=

0+1+4+9+ &c..

n.n-1

1 x 2

n.n-1.2n-1

1x2x3

PROB. 8. To find the sum of the infinite series 1+

2'

Whence 1=

4 5

and therefore s=2, the sum required.

PROB. 9. To find the sum of n terms of the above series.

PROB. 10. To find the sum s of the infinite series

Then will z=1— x.x + x2 + x3 + x*+x +,&c. which quantity, by actual multiplication, comes out =x, that is, x=z; and therefore, substituting x for z in the second step, it becomes x+x2+x3

+x++x5=- =s; in which, by restoring the value of x, we

-x

1 1 1 1 1

have + + + + +, &c. (

2 4 8 16 32

quired.

=

-)=1=s, the sum re

PROB. 11. To find the sum of 1000 terms of the series 1+ 5+9+13+17+, &c. Ans. 1999000.

PROB. 12. To find the sum of 20 terms of the series 1+3+ 9+27+81+, &c. Ans. 174339220.

PROB. 13. To find the sum of 12 terms of the series 4+9+ 16+25+, &c. Ans. 1562.

PROB. 14. To find the sum of n terms of the series a3+a+dl3 n.n- - 1.3 a'd

+a+2d+a+3 d3 +, &c. Ans. na3 +

2

+

PROB. 15. To find the sum of n terms of the series 1+3+

PROB. 16. Required the sum of the infinite series

PROB. 17. To find the sum of the infinite series

PROB. 19. To find the sum of the infinite series

+ 1.2.3 2.3.4

PROB. 20. To find the sum of n terms of the above series.

1

PROB. 21. To find the sum of the infinite series

PROB. 22. To find the sum of n terms of the above series.

Let there be given d=N, in which expression x is the logarithm of a*; it is required to find the value of x, that is, the logarithm of (a) the number N.

Let a=1+b, and N=1+n; then will 1+b=1+n, from which, extracting the yth root, we obtain 1+b=1+n}, ·.·

x

Here, if y be assumed indefinitely great, the quantities

y

y

-, may be considered as o, since they will in that case be inde

finitely small with respect to the numbers 1, 2, 3, 4, &c.

1

Wherefore -1=-1, -1=-1, -2=-2,

-2, &c.

y

y

y

1

2=

y

These values being substituted in the above series, we shall

either of the two latter fractions, then the last but one will be

1

come x (or the log. of 1+n) = ;·Ñ — — n3 + n 3 — n*+, &c. which

M

series, when n is a whole number, does not converge, and therefore is of no use; but we may obtain by means of it a series which will converge sufficiently fast for our purpose, as follows:

for n let -n be substituted, and the above expression becomes

And if the lower equation be subtracted from

the remainder is (log. 1+n-log. 1-n=) log.

the upper,

1+ n 1

1-n M.

be substituted for n in this equation, and it will become