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(B) Let the three consecutive odd numbers be $(2k-1)$,$(2k+1)$,and $(2k+3)$.The sum of their squares is $S = (2k-1)^2 + (2k+1)^2 + (2k+3)^2$.Expanding

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$12$ but not $24$

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(B) Let the three consecutive odd numbers be $(2k-1)$,$(2k+1)$,and $(2k+3)$.The sum of their squares is $S = (2k-1)^2 + (2k+1)^2 + (2k+3)^2$.Expanding these terms: $S = (4k^2 - 4k + 1) + (4k^2 + 4k + 1) + (4k^2 + 12k + 9) = 12k^2 + 12k + 11$.Increasing this sum by $1$,we get $S + 1 = 12k^2 + 12k + 12 = 12(k^2 + k + 1)$.Since $k^2 + k = k(k+1)$ is always even (as it is the product of two consecutive integers),let $k(k+1) = 2m$.Then $S + 1 = 12(2m + 1) = 24m + 12$.This expression is clearly divisible by $12$,but not necessarily by $24$ (since $12$ is not divisible by $24$).Thus,the sum is divisible by $12$ but not by $24$.

The sum of the squares of three consecutive odd numbers increased by $1$ is divisible by-

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