[SQRT(1)]+[SQRT(2)]+[SQRT(3)]+.......+[SQRT(2115)]=?
WHERE SQRT MEANS SQUARE ROOT
SAQUIB ASRAR , 11 Years ago
Grade 12
Algebra> [SQRT(1)]+[SQRT(2)]+[SQRT(3)]+.......+[SQ...
1 AnswersArun Kumar
if we see
[sqrt(1)]=1
[sqrt(2)]=1
in fact for
1 its 1*(2^2-1^2)
for 2 2*(3^2-2^2)
similarly [root(2115)]=45
so sigma n((n+1)^2-n^2) from 1 to 44
then we take t=45*(2115-2045+1)
(2n+1(n+1)(n)/3+(n+1)(n)/2 put n=44 then add with t
get the answer
[sqrt(1)]=1
[sqrt(2)]=1
in fact for
1 its 1*(2^2-1^2)
for 2 2*(3^2-2^2)
similarly [root(2115)]=45
so sigma n((n+1)^2-n^2) from 1 to 44
then we take t=45*(2115-2045+1)
(2n+1(n+1)(n)/3+(n+1)(n)/2 put n=44 then add with t
get the answer
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