Question

The sum of $(3^3-2^3)+(5^3-4^3)+(7^3-6^3)+.....to10$ brackets is

• A. $4960$
• B. $4860$
• C. $5060$
• D. $\text{None of these}$

Answer 1

The correct option is A $4960$

$1^3+2^3+3^3+4^3+...+{21}^3=\frac{n^2(n+1{)}^2}{4}$ ...(I)

using (1)

$\to 1^3+2^3+3^3+4^3+.........+{21}^3=\frac{(21{)}^2\times (22{)}^2}{4}=53361$ ...(II)

$1^3+3^3+5^3+7^3+.......+(2n-1{)}^3=n^2(2n^2-1)$ ...(III)

Using (III)

$1^3+3^3+5^3+....+(21{)}^3={11}^2(2\times {11}^2-1)=29161$ ...(IV)

Given series can be written as:-

$(1^3+3^3+5^3+7^3+9^3+{11}^3+{13}^3+{15}^3+{15}^3+{17}^3+{19}^3+{21}^3-1)$

$-(2^3+4^3+6^3+8^3+{10}^3+{12}^3+{14}^3+{16}^3+{18}^3+{20}^3)$ ...(V)

Now, on apply (II) - (IV), we get:-

$(2^3+4^3+6^3+...+{20}^3)=53361-29161$

$=24200$ ...(VI)

Now, Put the value of eqn (I), (VI) in (V), we get

$(29161-1)-\left(24200\right)$

$=4960$

$\therefore$ Hence, option A is correct.