The sum of the series 313-323-503 is

The correct option is C 1409400S = (1^3 + 2^3 + … +50^3) (1^3 + 2^3 + … +30^3) = ( 50 × 512 )^2 ( 30 × 312 )^2 [ Using ∑ n^3 = ( n(n+1)^22 ) ] =14(50 × ...

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Standard XII

Mathematics

Sigma n3

The sum of th...

Question

The sum of the series $31^{3} + 32^{3} + \dots \dots + 50^{3}$ is

98092

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121023

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150569

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1409400

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Solution

The correct option is D $1409400$
$S = (1^{3} + 2^{3} + \dots + 50^{3}) - (1^{3} + 2^{3} + \dots + 30^{3})$
$= {(\frac{50 \times 51}{2})}^{2} - {(\frac{30 \times 31}{2})}^{2}$
$[Using \sum n^{3} = (\frac{n (n + 1)^{2}}{2})]$
$= \frac{1}{4} (50 \times 51 - 30 \times 31) (50 \times 51 + 30 \times 31)$
$= 1409400$

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The sum of the series 313+323+……+503 is

The correct option is D 1409400S=(13+23+…+503)−(13+23+…+303) =(50×512)2−(30×312)2 [Using ∑n3=(n(n+1)22)] =14(50×51−30×31)(50×51+30×31) =1409400

The sum of the series $31^{3} + 32^{3} + \dots \dots + 50^{3}$ is