The correct option is A 42804Note that nbsp;9^3+10^3+11^3+...+20^3 =[1^3+2^3+...+20^3] [1^3+2^3+...+8^3] The sum of cubes of first n natural numbers is give ...

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Sum of Cubes of n Natural Numbers

93+103+113+…+...

Question

$9^{3} + 10^{3} + 11^{3} + . . . + 20^{3} =$ ____

42804

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44100

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45396

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42840

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Solution

The correct option is A 42804
Note that $9^{3} + 10^{3} + 11^{3} + . . . + 20^{3} = [1^{3} + 2^{3} + . . . + 20^{3}] - [1^{3} + 2^{3} + . . . + 8^{3}]$

The sum of cubes of first $n$ natural numbers is given by
$S_{n} = [\frac{n (n + 1)}{2}]^{2} .$
$\Rightarrow 1^{3} + 2^{3} + . . . + 20^{3} = [\frac{20 (20 + 1)}{2}]^{2} = 44100$
$\Rightarrow 1^{3} + 2^{3} + . . . + 8^{3} = [\frac{8 (8 + 1)}{2}]^{2} = 1296$

$∴ 9^{3} + 10^{3} + 11^{3} + . . . + 20^{3} = 44100 - 1296 = 42804$

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11^{3} - 10^{3} + 9^{3} - 8^{3} + 7^{3} - 6^{3} + 5^{3} - 2^{3} + 1^{3} =

11^{3} + 12^{3} + 13^{3} + . . . + 20^{3} =

97, 101, 103, 107, 111, 113

Q. Observe the following pattern:
1³ = 1
1³ + 2³ = (1 + 2)²
1³ + 2³ + 3³ = (1 + 2 + 3)²
Write the next three rows and calculate the value of 1³ + 2³+ 3³ + ... + 9³+ 10³ by the above pattern.

Q. The value of

∣ ∣ ∣ ∣ \begin{matrix} 103 & 115 & 114 111 & 108 & 106 104 & 113 & 116 \end{matrix} ∣ ∣ ∣ ∣ + ∣ ∣ ∣ ∣ \begin{matrix} 113 & 116 & 104 108 & 106 & 111 115 & 114 & 103 \end{matrix} ∣ ∣ ∣ ∣

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93+103+113+...+203= ____

The correct option is A 42804Note that 93+103+113+...+203=[13+23+...+203]−[13+23+...+83] The sum of cubes of first n natural numbers is given by Sn=[n(n+1)2]2. ⇒13+23+...+203=[20(20+1)2]2 =44100 ⇒13+23+...+83=[8(8+1)2]2 =1296 ∴93+103+113+...+203=44100−1296=42804

$9^{3} + 10^{3} + 11^{3} + . . . + 20^{3} =$ ____