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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'# Pixelmator 2 1 4 – powerful layer based image editor.
#• ' prove true for n = k + 1'#
#n=1toLHS=1^2=1#
#'and RHS ' =1/6(1+1)(2+1)=1#
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#rArrcolor(red)'result is true for n = 1'# https://torrent-bath.mystrikingly.com/blog/5-nights-at-freddys-night-1.
#'assume result is true for n = k'#
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#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#
#=1/6(k+1)[k(2k+1)+6(k+1)]# Atom editor online.
#=1/6(k+1)(2k^2+7k+6)#
#=1/6(k+1)(k+2)(2k+3)#
#=1/6n(n+1)(2n+1)to' with ' n=k+1#
#rArrcolor(red)'result is true for n = k + 1'#
Liquid flow pro 17. #rArr1^2+2^2+3^2+.+n^2=1/6n(n+1)(2n+1)#
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},#
#=(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,#
Speed read 2 0 1 – reading technique ielts. # :. 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,#
# 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,#
Speed read 2 0 1 – reading technique ielts. # :. 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,#
# :. 2n^3+3n^2+n=6S_n,#
# :. 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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