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Can you prove that #1^2+2^2+3^2+.+n^2=1/6n(n+1)(2n+1)#?

2 Answers

Explanation:

#'using the method of 'color(blue)'proof by induction'#

#'this involves the following steps '#

#• ' prove true for some value, say n = 1'#

#• ' assume the result is true for n = k'#

#• ' prove true for n = k + 1'#

#n=1toLHS=1^2=1#

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#'and RHS ' =1/6(1+1)(2+1)=1#

#rArrcolor(red)'result is true for n = 1'#

#'assume result is true for n = k'#

#color(magenta)'assume ' 1^2+2^2+ . +k^2=1/6k(k+1)(2k+1)#

#'prove true for n = k + 1'#

#1^2+2^2+.+k^2+(k+1)^2=1/6k(k+1)(2k+1)+(k+1)^2#

Adobe acrobat 9 gratuit. #=1/6(k+1)[k(2k+1)+6(k+1)]#

#=1/6(k+1)(2k^2+7k+6)#

#=1/6(k+1)(k+2)(2k+3)#

Permute 3 v3 1. #=1/6n(n+1)(2n+1)to' with ' n=k+1#

#rArrcolor(red)'result is true for n = k + 1'#

#rArr1^2+2^2+3^2+.+n^2=1/6n(n+1)(2n+1)#

#'prove true for n = k + 1'#

#1^2+2^2+.+k^2+(k+1)^2=1/6k(k+1)(2k+1)+(k+1)^2#

Adobe acrobat 9 gratuit. #=1/6(k+1)[k(2k+1)+6(k+1)]#

#=1/6(k+1)(2k^2+7k+6)#

#=1/6(k+1)(k+2)(2k+3)#

Permute 3 v3 1. #=1/6n(n+1)(2n+1)to' with ' n=k+1#

#rArrcolor(red)'result is true for n = k + 1'#

#rArr1^2+2^2+3^2+.+n^2=1/6n(n+1)(2n+1)#

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Explanation:

Let, #S_n=1^2+2^2+3^2+.+n^2, &, , f(n)=n^3, n in NNuu{0}.#

#:. f(n)-f(n-1)=n^3-(n-1)^3.#

#because, a^3-b^3=(a-b)(a^2+ab+b^2), f(n)-f(n-1),#

#={n-(n-1)}{n^2+n(n-1)+(n-1)^2},#

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#=(1)(n^2+n^2-n+n^2-2n+1),#

# rArr f(n)-f(n-1)=n^3-(n-1)^3=3n^2-3n+1;(n in NNuu{0}.#

#n=1 rArr 1^3-0^3=3(1)^2-3(1)+1;#
#n=2 rArr 2^3-1^3=3(2)^2-3(2)+1;#
#n=3 rArr 3^3-2^3=3(2)^2-3(2)+1;#
#vdots vdots vdots vdots vdots vdots vdots vdots vdots vdots#
#n=n rArr n^3-(n-1)^3=3(n)^2-3(n)+1;#

#'Adding, 'n^3-0^3=3{1^2+2^2+3^2+.+n^2}-3{1+2+3+.+n}+n,#

# :. n^3=3S_n-3Sigman+n, or, #

# n^3=3S_n-3/2n(n+1)+n, i.e.,#

#2n^3=6S_n-3n(n+1)+2n=6S_n-3n^2-3n+2n,#

Cakewalk z3ta 2. T software download. # :. 2n^3+3n^2+n=6S_n,# https://amranbiane1982.mystrikingly.com/blog/beta-1-5-minecraft-download.

# :. 6S_n=n(2n^2+3n+1)=n(n+1)(2n+1),#

# rArr S_n=n/6(n+1)(2n+1).#

Enjoy Maths.!

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