# Pillow-Problems: Problem #21

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Problem #19 | Problem #22
For more information on this collection, see Pillow-Problems by Charles L. Dodgson (Lewis Carroll).

## Problem:

21.

Sum (1) to n terms, (2) to 100 terms, the series

1 · 3 · 5; + 2 · 4 · 6; + &c.
[7/4/89

21.

(1)
 n · n + 1 · n + 4 · n + 5 4
; (2) 27,573,000.

## Solution:

21.

(1) The nth term is n · n + 2 · n + 4;

∴the (n + 1)th term is n + 1 · n + 3 · n + 5;
= (n + 1) ·(n + 2 + 1) · (n + 5);
= n + 1 · n + 2 · n + 5 · n + 1 · n + 5
= n + 1 · n + 2 · (n + 3 + 2) + n + 1 · (n + 2 + 3);
= n + 1 · n + 2 · n + 3 + 2 · n + 1 · n + 2 + n + 1 · n + 2 + 3 · n + 1;
= n + 1 · n + 2 · n + 3 + 3 · n + 1 · n + 2 + 3 · n + 1.
S =
 n · n + 1 · n + 2 · n + 3 4
+ n · n + 1 · n + 2 +
 (3)/ (2)
· n · n + 1 + C;
and C = 0.
S = n · n + 1 · (
 n² + 5n + 6 4
+ n + 2 +
 (3)/ (2)
);
= n · n + 1 ·
 n² + 9n + 20 4
=
 n · n + 1 · n + 4 · n + 5 4

Q.E.F.

(2) S, to 100 terms,

=
 100 · 101 · 104 · 105 4
= 100 · 101 · 26 · 105;
now 101 · 105 = 10,605;
∴ 101 · 105 · 13 = 130,000 + 7800 + 65 = 137,865;
and twice this = 27400 + 1730 = 275,730;
S = 27,573,000.

Q.E.F.