Question

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How many terms of the series 13 + 23 + 33 + … should be taken to get the sum 14400?​

Answer 1

$\sf{ {1}^{3} + {2}^{3} + {3}^{3} + _\dots + {n}^{3} = 14400}$

$= \sf{ \bigg( \dfrac{n(n + 1)}{2} \bigg)^{2} } = 14400 = (120)^{2}$

$= \sf{ \dfrac{n(n + 1)}{2} } = \sqrt{14400 }= 120$

$\sf{ = n(n + 1) = 240}$

Method 1:-

$\sf{ = {n}^{2} + n - 240 = 0}$

$\sf{ {n}^{2} + 16n - 15n - 240 = 0}$

$= \sf{n(n + 16) - 15(n + 16) = 0}$

$\sf{(n + 16)(n - 15) = 0}$

$\sf{n + 16 = 0}$

$\sf{n = - 16}$

$\sf{n - 15 = 0}$

$\sf{n = 15}$

$\bigstar { \underline{\boxed{\textsf{∴ 15 terms to be taken to get the sum 14400.} }}} \bigstar$

Method 2:-

$\sf{ {n}^{2} + n - 240 = 0}$

$\sf{n = \dfrac{ - b ± \sqrt{ {b}^{2} - 4ac} }{2a} }$

$\sf{ = \dfrac{ - 1±\sqrt{1 -4 \times 1 + ( - 240) } }{2 \times 1}}$

$\sf{ = \dfrac{ - 1 \: ± \: \sqrt{961} }{2}}$

$\sf{ = \dfrac{ - 1 \: ± \: 31}{2}}$

$\sf{ = \dfrac{ - 1 + 31}{2}} \: or \: \dfrac{ - 1 - 31}{2}$

$= \sf{\dfrac{30}{2} \: or \:\dfrac{ - 32 }{2}}$

$= \sf{ \cancel\dfrac{30}{2} \: or \: \cancel\dfrac{ - 32 }{2}}$

$\sf{n = 15 \: or \: n = - 16}$

$\sf{n \: cannot \: be = - 16}$

$\dashrightarrow \underline{ \boxed{ \therefore \frak \pink{n = 15}}}\bigstar$

Answer 2

Answer :—

15th term .

Solution :—

$\sf{ = {n}^{2} + n - 240 = 0}$

$= \sf{ {n}^{2} + 16n - 15n - 240 = 0}$

$= \sf{n(n + 16) - 15(n + 16) = 0}$

$= \sf{(n + 16)(n - 15) = 0}$

$= \sf{n + 16 = 0}$

$= \sf{n = - 16}$

$= \sf{n - 15 = 0}$

$= \sf{n = 15}$

Therefore , 15 terms to be taken to get the sum 14400.