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Question

$\frac{1^{3} + 2^{3} + 3^{3} + 4^{3} + . . . . . . . {.12}^{3}}{1^{2} + 2^{2} + 3^{2} + 4^{2} + . . . . . {.12}^{2}} =$

\frac{234}{25}

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\frac{243}{35}

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\frac{263}{27}

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None of these

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Solution

The correct option is A $\frac{234}{25}$
Clearly required ratio is $= \frac{\sum n^{3}}{\sum n^{2}} = \frac{\frac{n^{2} (n + 1)^{2}}{4}}{\frac{n (n + 1) (2 n + 1)}{6}} = \frac{3 n (n + 1)}{2 (2 n + 1)}$
Substituting $n = 12$ we get required value $= \frac{234}{25}$
Hence, option 'A' is correct.

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$C I$	$8 - 12$	$13 - 17$	$18 - 22$	$23 - 27$	$28 - 32$	$33 - 37$	$38 - 42$	$43 - 47$
$f$	$14$	$8$	$20$	$7$	$11$	$10$	$24$	$6$

42

50

29, 45, 23, 40, 31, 11, 48, 1, 30, 24, 25, 29, 25, 32, 31, 22, 9, 49, 19, 13, 32, 39, 25, 3, 27, 41,

12, 13, 2, 44, 7, 43, 7, 43, 15, 35, 40, 3, 12, 48, 49, 18

\begin{matrix} S u b j e c t s A n i l & I & I I & I I I & I V & V M a t h s & 42 & 32 & 47 & 22 & 37 D I & 32 & 42 & 27 & 32 & 22 V A & 37 & 27 & 17 & 32 & 42 R e a s o n i n g & 47 & 49 & 34 & 41 & 39 E n g l i s h & 26 & 30 & 38 & 41 & 45 B a r u n M a t h s & 32 & 22 & 37 & 47 & 42 D I & 27 & 47 & 32 & 39 & 30 V A & 35 & 29 & 42 & 36 & 28 R e a s o n i n g & 44 & 45 & 42 & 45 & 47 E n g l i s h & 32 & 30 & 39 & 36 & 33 C h a m a n M a t h s & 33 & 34 & 37 & 44 & 39 D I & 44 & 42 & 35 & 42 & 38 V A & 36 & 43 & 42 & 40 & 39 R e a s o n i n g & 26 & 37 & 46 & 34 & 35 E n g l i s h & 27 & 29 & 35 & 35 & 43 D h a n n u M a t h s & 29 & 36 & 37 & 42 & 43 D I & 33 & 39 & 37 & 29 & 31 V A & 44 & 45 & 39 & 49 & 45 R e a s o n i n g & 31 & 36 & 48 & 43 & 43 E n g l i s h & 35 & 38 & 36 & 41 & 45 E l l a M a t h s & 35 & 34 & 43 & 45 & 46 D I & 47 & 33 & 44 & 39 & 47 V A & 47 & 44 & 39 & 43 & 42 R e a s o n i n g & 39 & 37 & 36 & 42 & 43 E n g l i s h & 32 & 33 & 38 & 41 & 44 \end{matrix}

\frac{3 \times 1^{3}}{1^{2}} + \frac{5 \times (1^{3} + 2^{3})}{1^{2} + 2^{2}} + \frac{7 \times (1^{3} + 2^{3} + 3^{3})}{1^{2} + 2^{2} + 3^{2}} + \dots

10^{th}

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