find the smallest integer n>1 such that the sum 1+2+…+n is a perfect square

1 thought on “find the smallest integer n>1 such that the sum 1+2+…+n is a perfect square”

1. [tex]\textbf{Given:}[/tex]

   [tex]\textsf{1+2+3+. . . . . . +n is a perfect square}[/tex]

   [tex]\textbf{To find:}[/tex]

   [tex]\textsf{The smallest integer satisfying the given condition}[/tex]

   [tex]\textbf{Solution:}[/tex]

   [tex]\textsf{Consider,}[/tex]

   [tex]\mathsf{1+2+\;.\;.\;.\;.+n}[/tex]

   [tex]\mathsf{=\dfrac{n(n+1)}{2}}[/tex]

   [tex]\mathsf{But,\;as\;per\;given\;data,}[/tex]

   [tex]\mathsf{\dfrac{n(n+1)}{2}\;is\;a\;perfect\;square}[/tex]

   [tex]\mathsf{For\;n=2,\;\dfrac{2{\times}3}{2}=3}[/tex]

   [tex]\mathsf{For\;n=3,\;\dfrac{3{\times}4}{2}=6}[/tex]

   [tex]\mathsf{For\;n=4,\;\dfrac{4{\times}5}{2}=10}[/tex]

   [tex]\mathsf{For\;n=5,\;\dfrac{5{\times}6}{2}=15}[/tex]

   [tex]\mathsf{For\;n=6,\;\dfrac{6{\times}7}{2}=21}[/tex]

   [tex]\mathsf{For\;n=7,\;\dfrac{7{\times}8}{2}=28}[/tex]

   [tex]\mathsf{For\;n=8,\;\dfrac{8{\times}9}{2}=36,\;a\;perfect\;square}[/tex]

   [tex]\therefore\underline{\textsf{The smallest value of n satisfying the given condition is 8}}[/tex]

   [tex]\textbf{Find more:}[/tex]

   The smallest value of x that satisfies the equation 2^2x – 8 * 2^x = -12 is ?

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