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Example 1. Find the 7th term and the sum of the first seven terms of the series 4, 14, 30, 52, 80,... The successive orders of differences are

10, 16, 22, 28,
6, 6, 6,

0, 0.
Here n=7, and a=4.
Hence, using formula, Art. 494, the 7th term

6.5
=4+6.10+ .6=154.

1.2
Using formula, Art. 496, the sum of the first seven terms

7.6 7.6.5 =7.4+

.6=448.

1.2 1.2.3 Example 2. Find the general term and the sum of n terms of the series

12, 40, 90, 168, 280, 432, .... The successive orders of difference are

28, 50, 78, 112, 152, ......
22, 28, 34,

40, ......
6, 6, 6,...
0,

0,... Hence the nth term [Art. 494]

22(n-1)(n-2), 6(n-1)(n-2)(n-3) =12+28(n-1)+

2

13 =n8+5n2+6n. Using the formula for the sum of n terms we obtain Sn=12n+ 28n(n-1) 22n(n-1)(n-2) on(n-1)(n-2) (11-3) +

+
12
13

14
(3n2+26n+69n+46)
12
1

n(n+1) (3n2+23n+46).
12

n =

497. It will be seen that this method of summation will only succeed when the series is such that in forming the orders of differences we eventually come to a series in which all the terms are equal. This will always be the case if the nth term of the series is a rational integral function of n.

PILES OF SHOT AND SHELLS.

498. To find the number of shot arranged in a complete pyramid on a square base.

The top layer consists of a single shot; the next contains 4; the next 9, and so on to na, n being the number of layers: hence the form of the series is 12, 22, 3?,

na. Series

1,

16, 1st order of differences 2nd

2,

2, 3rd

0. Substituting in Art. 496, we obtain

n(n-1) n)-) n(n+1) (2n+1)
S=nt
1.2
1.2.3

6

4,

9,

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3,

5,

7,

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66

..3+

3,

4,

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499. To find the number of shot arranged in a complete pyramid the base of which is an equilateral triangle.

The top layer consists of a single shot; the next contains 3; the next 6; the next 10, and so on, giving a series of the form

1, 1+2, 1+2+3, 1+2+3+4, ...... Series 1, 3, 6,

10. 1st order of differences 2, 2nd

1, 1, 3rd

0.

n(n-1)(n-2) n(n+1)(n+2) Hence S=nt

1.2
1.2.3

6 500. To find the number of shot arranged in a complete pile the base of which is a rectangle.

The top layer consists of a single row of shot. Suppose this row to contain m shot; then the next layer contains 2(m+1); the next 3(m+2), and so on, giving a series of the form

m,

2m +2, 3m +6, 4m +12, ... 1st order of differences m+2, m +4, m+6, 2nd

2,

2, 3rd

0.

n(n-1).2+

1=

66

Now let l and w be the number of shot in the length and width, respectively, of the base; then m=l-w+1. Making these substitutions, we have

n(n+1) (31-1+1) S=

6

EXAMPLES XLVI. b.

1. Find the eighth term and the sum of the first eight terms of the series 1, 8, 27, 64, 125, ..

2. Find the tenth term and the sum of the first ten terms of the series 4, 11, 28, 55, 92, ...

Find the number of shot in : 3. A square pile, having 15 shot in each side of the base. 4. A triangular pile, having 18 shot in each side of the base.

5. A rectangular pile, the length and the breadth of the base containing 50 and 28 shot respectively.

6. An incomplete triangular pile, a side of the base having 25 shot, and a side of the top 14.

7. An incomplete square pile of 27 courses, having 40 shot in each side of the base.

8. Find the ninth term and the sum of the first nine terms of the series 1, 3+5, 7+9+11, ...

The numbers 1, 2, 3, ... are often referred to as the natural numbers.

9. Find the sum of the squares of the first n natural numbers. 10. Find the sum of the cubes of the first n natural numbers.

11. The number of shot in a complete rectangular pile is 24395; if there are 34 shot in the breadth of the base, how many are there in its length?

12. The number of shot in the top layer of a square pile is 169, and in the lowest layer is 1089; how many shot does the pile contain ?

13. Find the number of shot in a complete rectangular pile of 15 courses, having 20 shot in the longer side of its base.

14. Find the number of shot in an incomplete rectangular pile, the number of shot in the sides of its upper course being 11 and 18, and the number in the shorter side of its lowest course being 30.

Find the nth term and the sum of n terms of the series :

15. 4, 14, 30, 52, 80, 114, ...
16. 8, 26, 54, 92, 140, 198, ...
17. 2, 12, 36, 80, 150, 252, ...
18. 8, 16, 0, -64, – 200, – 432, ...
19. 30, 144, 420, 960, 1890, 3360, ...

20. What is the number of shot required to complete a rectangular pile having 15 and 6 shot in the longer and shorter side, respectively, of its upper course ?

21. The number of shot in a triangular pile is greater by 150 than half the number of shot in a square pile, the number of layers in each being the same ; find the number of shot in the lowest layer of the triangular pile.

22. Find the number of shot in an incomplete square pile of 16 courses when the number of shot in the upper course is 1005 less than in the lowest course.

23. Show that the number of shot in a square pile is onefourth the number of shot in a triangular pile of double the number of courses.

24. If the number of shot in a triangular pile is to the number of shot in a square pile of double the number of courses as 13 to 175, find the number of shot in each pile.

INTERPOLATION.

501. The process of introducing between the terms of a series intermediate values conforming to the law of the series is called interpolation. An important application is in finding numbers intermediate between those given in Logarithmic and other mathematical tables. For this purpose we may employ the formula used in finding the nth term by the Differential Method, giving fractional values to n.

Example. Given log 40=1.6021, log 41=1.6128, log 42=1.6232, log 43=1.6335, find log 40:7. Series 1.6021, 1.6128, 1.6232,

1•6335, 1st order of differences •0107, 0104, 2nd

- •0003, —.0001, 3rd

+.0002.

·0103,

66

66

Substituting in formula of Art. 494, we have

7

7 3 .0003 log 40:7=1.6021+ (•0107) +

10

10 10 | 2
7 3 13 .0002
+

10 10 10 | 3
=1.6021 +.00749+.000031 +.000009

=1.6096+. Here log 40 is the first term (n=1); log 41 is the second term (n=2); hence in introducing the intermediate term log 40•7 we give to n a value 1.7.

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EXAMPLES XLVI. C.

......

1. Given log 3=0.4771, log 4=0.6021, log 5=0.6990, log 6 =07782, ... ; find log 4.4.

2. Given log 51=17076, log 52=1.7160, log 53=1.7243, log 54 =1.7324, ; find log 51.9.

3. Given V5=2.236, V6=2.449, V7=2.645, V8=2.828 ; find V5:6, 77.4, and 77:74.

4. Given 351–3.7084, 352=3•7325, 353=3.7563, ...; find V51.18.

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