How can I write the statement #S_1, S_2, S_3# for this?(see picture) Thanks a lot!

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Answer 1 · 2018-03-15T06:38:06

Please see below.

We are given #S_n=1^2+4^2+7^2+...+(3n-2)^2=(n(6n^2-3n-1))/2#

For #S_1#, while LHS is #1^2=1#, RHS is #(1(6*1^2-3*1-1))/2=1#.

Hence for #n=1# it is true.

For #S_2#, while LHS is #1^2+4^2=1+16=17#, RHS is #(2(6*2^2-3*2-1))/2=(2(24-6-1))/2=(2*17)/2=17#.

Hence for #n=2# it is also true.

For #S_3#, while LHS is #1^2+4^2+7^2=1+16+49=66#, RHS is #(3(6*3^2-3*3-1))/2=(3(54-9-1))/2=(3*44)/2=66#.

Hence for #n=3# it is also true.