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Let the numbers be m & m+1 (consecutive) Hence their squares are m^2 & ( m + 1)^2 => m^2 & m^2 + 2m + 2 Hence Their sum is m^2 + m^2 + 2m + 1 = 85 2m^2 + 2m - 84 = 0 m^2 + m - 42 =0 Factor (m + 7)(m - 6) = 0 Hence the numbers are 6 & 7 .
The equation for n layers is S(n) = n(n+1)(2n+1)/6It is simplest to prove it by induction.When n = 1,S(1) = 1*(1+1)(2*1+1)/6 = 1*2*3/6 = 1.Thus the formula is true for n = 1.Suppose it is true for n = m. That is, for a pyramid of m levels,S(m) = m*(m+1)*(2m+1)/6Then the (m+1)th level has (m+1)*(m+1) oranges and soS(m+1) = S(m) + (m+1)*(m+1)= m*(m+1)*(2m+1)/6 + (m+1)*(m+1)= (m+1)/6*[m*(2m+1) + 6(m+1)]= (m+1)/6*[2m^2 + m + 6m + 6]= (m+1)/6*[2m^2 + 7m + 6]= (m+1)/6*(m+2)*(2m+3)= (m+1)*(m+2)*(2m+3)/6= [(m+1)]*[(m+1)+1)]*[2*(m+1)+1]/6Thus, if the formula is true for n = m, then it is true for n = m+1.Therefore, since it is true for n =1 it is true for all positive integers.
(m-2)/(m+2) * m/(m-1) = [(m-2)*m]/[(m+2)*(m-1)] = (m2 - 2m)/m2 + m - 2)
m^2 6n - 9n2 - 2m = (m - 3n)(m + 3n - 2)
Let the integers be 'm' & 'm+`1'. Hence m + m+ 1 = 57 2m + 1 = 57 2m = 56 m = 28 M+ 1 = 28 + 1 = 29