Pertanyaan

Diketahui i =1 n i 2 = 1 2 + 2 2 + 3 2 +…+ n 2 = n n +1 2 n +1 6 benar untuk n = k + 1 n = k + 1 . Pernyataan diatas dapat ditulis sebagai …

Diketahui $sum subscript i equals 1 end subscript superscript n i squared equals 1 squared plus 2 squared plus 3 squared plus horizontal ellipsis plus n squared equals fraction numerator n open parentheses n plus 1 close parentheses open parentheses 2 n plus 1 close parentheses over denominator 6 end fraction$ benar untuk $n = k + 1$ . Pernyataan diatas dapat ditulis sebagai …

$1 squared plus 2 squared plus 3 squared plus horizontal ellipsis plus open parentheses k plus 1 close parentheses squared equals fraction numerator open parentheses k plus 1 close parentheses squared open parentheses 2 k plus 3 close parentheses over denominator 6 end fraction$
$1 squared plus 2 squared plus 3 squared plus horizontal ellipsis plus open parentheses k plus 1 close parentheses squared equals fraction numerator open parentheses k plus 1 close parentheses open parentheses k plus 2 close parentheses open parentheses k plus 3 close parentheses over denominator 6 end fraction$
$1 squared plus 2 squared plus 3 squared plus horizontal ellipsis plus open parentheses k plus 1 close parentheses squared equals fraction numerator open parentheses k plus 1 close parentheses open parentheses k plus 2 close parentheses open parentheses 2 k plus 3 close parentheses over denominator 6 end fraction$
$1 squared plus 2 squared plus 3 squared plus horizontal ellipsis plus k squared equals fraction numerator open parentheses k close parentheses open parentheses k plus 1 close parentheses open parentheses 2 k plus 1 close parentheses over denominator 6 end fraction$
$1 squared plus 2 squared plus 3 squared plus horizontal ellipsis plus open parentheses k plus 1 close parentheses squared equals fraction numerator open parentheses k plus 1 close parentheses open parentheses k plus 1 close parentheses open parentheses 2 k plus 1 close parentheses over denominator 6 end fraction$