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Sums of Powers of Natural Numbers

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Sums of Powers of Natural Numbers Sums of Powers of Natural Numbers >2 We'll use the symbol W55 for the sum of the powers of the first 8 natural numbers. In other words, 55 5 Wœ"#ÞÞÞ8Þ5 Of course, this is a “formula” for W5, but it doesn't help you compute it doesn't tell you how to find the $$ $ exact value, say, of Wœ"#ÞÞÞ"&$ . We'd like to get what's called a closed formula for W5, that is, one without the annoying “ ... ” in it. !! ! For Wœ"#ÞÞÞ8! , this is easy: since there are 8 terms, each equal to ", so we get W! œ""ÞÞÞ" œ"†8 œ8 For WW", it's already harder. Here's a slick way of finding a closed formula for ": Write down W" twice, in two different orders: W" œ " # $ ÞÞÞ Ð8 "Ñ 8, and W" œ 8 Ð8 "Ñ Ð8 #Ñ ÞÞÞ # " Then add to get: __________________________________________________ #W" œ Ð8 "Ñ Ð8 "Ñ Ð8 "Ñ ... + Ð8 "Ñ Ð8 "Ñ. Since there are 8Ð8"Ñß terms on the right, each equal to we get #W" œ 8Ð8 "Ñ, so 8Ð8"Ñ Wœ" # "&Ð"'Ñ This is a “usable” closed formula: for example, "#$ÞÞÞ"&œ# œ"#!. Here's a list of formulas for: !! ! Wœ"#ÞÞÞ8œ8! "" "8Ð8"Ñ Wœ"#ÞÞÞ8œ" # ## #8Ð8"ÑÐ#8"Ñ Wœ"#ÞÞÞ8œ# ' 8Ð8"Ñ ## Wœ$$"[# ] (Curious observation: WœW[ ] ) Where do these formulas come from? There is a systematic way to get a formula for each W5 once you know the previous formulas for W!" ß W ß ÞÞÞß W 5" À We'll illustrate here with just two examples: !! ! i) If someone noticed the (easy) fact the W! œ" # ÞÞÞ8 œ""ÞÞÞ"œ8 "" " how could this fact be used to get a formula: W" œ" # ÞÞÞ8 œ "#ÞÞÞ8œ ??? Well, for any positive integer 4, we know that Ð4"Ñ## 4 œ#4". So we write this down substituting in each value 4 œ "ß 4 œ #ß ÞÞÞß 4 œ 8 ### " œ #Ð"Ñ " $## # œ #Ð#Ñ " %## $ œ #Ð$Ñ " ã Ð8"Ñ## 8 œ#Ð8Ñ" Adding up the columns on both sides (with lots of cancellations on the left-hand side) gives _____________________ Ð8"Ñ# " œ#Ð"#ÞÞÞ8Ñ8 œ# W" 8 so # 8 #8""œ#W" 8 so # 88œ#W" so 88# 8Ð8"Ñ ##œ œWÞ" ii) So now we know formulas for WW!" and . How can use these to we get a formula for ## # W# œ"#ÞÞÞ8= ??? It's the same idea, but just a little more algebra. For any 4 we know that Ð4 "Ñ$$ 4 œ $4 # $4 ". So we write this down substituting each value 4 œ "ß 4 œ #ß ÞÞÞß 4 œ 8 #"$$ œ$Ð"Ñ$Ð"Ñ" # $#$$ œ$Ð#Ñ$Ð#Ñ" # %$$$ œ$Ð$Ñ$Ð$Ñ" # ã Ð8"Ñ$$ 8 œ$Ð8Ñ$Ð8Ñ" # . Adding up the columns on both sides (with lots of cancellations on the left-hand side) gives ____________________________ Ð8"Ñ$### " œ$Ð" # ÞÞÞ8 Ñ$Ð"#ÞÞÞ8ÑÐ""ÞÞ"Ñ ÆÆÆ œ$ W#"! $ W W ß so $# 8 $8 $8 " " œ $W#"! $W W. Then we solve to get W #. $# 8 $8 $8 $W"! W œ $W #, so $# 8 $8 $8$W"! W Wœ# $ . We can substitute the formulas we already know for WW!" and and simplfy to ge $# 8Ð8"Ñ 8 $8 $8 $[]# 8 8Ð8"ÑÐ#8"Ñ Wœ# $'œœœ... Optional Material If you know the binomial formula (from high school) and can therefore expand Ð4 "Ñ5, then the same idea works for any natural number 55. But the bigger is, the more lagebra is involved. An outline goes like this. The formula for the “binomial coefficients” : 5 œÑ5x 66xÐ56Ñx Suppose we have figured out formulas for WßWßWßÞÞÞßW!"# 5". We know (from the binomial theorem) that for any 4, Ð4"Ñ4œ5" 5"5" 4 5 5" 4 5" 5" 4ÞÞÞ" 5# "# $ Write this out for each value 4 œ "ß #ß ÞÞÞ8. 25" " 5" œ5" " 5 5" " 5" 5" " 5# ÞÞÞ " "# $ $5" # 5" œ5" # 5 5" # 5" 5" # 5# ÞÞÞ " "# $ ÞÞÞÞÞÞ Ð8 "Ñ5" 8 5" œ5" 8 5 5" 8 5" 5" 8 5# ÞÞÞ ". Add the columns: "# $ __________________________________________________________ Ð8"Ñ5" 1 œ5" Ð" 5 # 5 ÞÞÞ8 5 Ñ 5" Ð" 5" # 5" ÞÞÞ8 5" Ñ "# 5" Ð"5# # 5# ÞÞÞ 8 5" Ñ ÞÞÞ 8 $ œWWWÞÞÞW5" 5" 5" . "#55"5#! $ Then we solve for what we want: WœÐ8"Ñ"W[]5" 5" 5" WÞÞÞWÎ 5" 55"5#!#$ " œÐ8"Ñ"WWÞÞÞWÎÐ5"Ñ[]5" 5" 5" #$5" 5# ! We are assuming that we already have formulas for WßWßÞÞÞWW5" 5# ", ! which we then substitute into this formula to get one closed, if complicated, formula for W8Þ5 in terms of Try it to find a formula for %% % Wœ"#ÞÞÞ8œ% ....
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