# formula 6

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\end{aligned}Sn​Sn​​==​1n​++​2n−1​++​3n−2​+⋯++⋯+​n1.​, Grouping and adding the above two sums gives, 2Sn=(1+n)+(2+n−1)+(3+n−2)+⋯+(n+1)=(n+1)+(n+1)+(n+1)+⋯+(n+1)⏟n times=n(n+1).\begin{aligned}
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The sum of the first nnn even integers is 222 times the sum of the first nnn integers, so putting this all together gives. Here sa,ns_{a,n}sa,n​ is the sum of the first nnn atha^\text{th}ath powers. This gives, n3=3(∑k=1nk2)−3∑k=1nk+∑k=1n1n3=3(∑k=1nk2)−3n(n+1)2+n3(∑k=1nk2)=n3+3n(n+1)2−n⇒∑k=1nk2=13n3+12n2+16n=n(n+1)(2n+1)6.\begin{aligned}

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