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14. Suppose you begin with a pile of n stones and split this pile into n piles of one stone each by successively splitting a pile of stones into two smaller piles. Each time you split a pile you multiply the number of stones in each of the two smaller piles you form, so that if these piles haver and s stones in them, respectively, you compute rs. Show that no matter how you split the piles, the sum of the products computed at each step equals n (n-1)/2.

14. Suppose you begin with a pile of n stones and split this pile into n piles of one stone each by successively splitting a pile of stones into two smaller piles. Each time you split a pile you multiply the number of stones in each of the two smaller piles you form, so that if these piles haver and s stones in them, respectively, you compute rs. Show that no matter how you split the piles, the sum of the products computed at each step equals n (n-1)/2.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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14. Suppose you begin with a pile of \( n \) stones and split this pile into \( n \) piles of one stone each by successively splitting a pile of stones into two smaller piles. Each time you split a pile, you multiply the number of stones in each of the two smaller piles you form, so that if these piles have \( r \) and \( s \) stones in them, respectively, you compute \( rs \). Show that no matter how you split the piles, the sum of the products computed at each step equals \( n(n - 1)/2 \).
Transcribed Image Text:14. Suppose you begin with a pile of \( n \) stones and split this pile into \( n \) piles of one stone each by successively splitting a pile of stones into two smaller piles. Each time you split a pile, you multiply the number of stones in each of the two smaller piles you form, so that if these piles have \( r \) and \( s \) stones in them, respectively, you compute \( rs \). Show that no matter how you split the piles, the sum of the products computed at each step equals \( n(n - 1)/2 \).
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