Question

Find the sigma notation of the following sum: $[(\frac{2}{n})^3-\frac{2}{n} ](\frac{2}{n} )+...+[(\frac{2n}{n} )^3-\frac{2n}{n} ](\frac{2}{n} )$​

Answer 1

Basic concept used :-

How to represent in Summation form

A series can be written in an abbreviated form by using the Greek letter ∑(sigma), called the summation sign.

• The variable of summation is represented by an index which is placed beneath the summation sign.

• The index is often represented by i. (Other common possibilities for representation of the index are j and t.)

• The index appears as the expression i = 1. The index assumes values starting with the value on the right hand side of the equation and ending with the value above the summation sign.

• The starting point for the summation or the lower limit of the summation

• The stopping point for the summation or the upper limit of summation

$\large\underline{\sf{Solution-}}$

$\large\underline{\sf{Given \: that- }}$

$\sf \: \bigg[ {\bigg( \dfrac{2}{n} \bigg) }^{3} - \dfrac{2}{n} \bigg]\dfrac{2}{n} + - - - + \bigg[ {\bigg( \dfrac{2n}{n} \bigg) }^{3} - \dfrac{2n}{n} \bigg]\dfrac{2}{n}$

can be re- written as

$= \sf \: \bigg[ {\bigg( \dfrac{2 \times 1}{n} \bigg) }^{3} - \dfrac{2 \times 1}{n} \bigg]\dfrac{2}{n} + \bigg[ {\bigg( \dfrac{2 \times 2}{n} \bigg) }^{3} - \dfrac{2 \times 2}{n} \bigg]\dfrac{2}{n} + - - - + \bigg[ {\bigg( \dfrac{2 \times n}{n} \bigg) }^{3} - \dfrac{2 \times n}{n} \bigg]\dfrac{2}{n}$

can be written in summation form as

$\sf \: = \displaystyle\:\sum_{i=1}^n \sf \: \bigg[ {\bigg( \dfrac{2i}{n} \bigg) }^{3} - \dfrac{2i}{n} \bigg]\dfrac{2}{n}$

Answer 2

Basic concept used :-

How to represent in Summation form

A series can be written in an abbreviated form by using the Greek letter ∑(sigma), called the summation sign.

The variable of summation is represented by an index which is placed beneath the summation sign.

The index is often represented by i. (Other common possibilities for representation of the index are j and t.)

The index appears as the expression i = 1. The index assumes values starting with the value on the right hand side of the equation and ending with the value above the summation sign.

The starting point for the summation or the lower limit of the summation

The stopping point for the summation or the upper limit of summation

\large\underline{\sf{Solution-}}

Solution−

\large\underline{\sf{Given \: that- }}

Giventhat−

\sf \: \bigg[ {\bigg( \dfrac{2}{n} \bigg) }^{3} - \dfrac{2}{n} \bigg]\dfrac{2}{n} + - - - + \bigg[ {\bigg( \dfrac{2n}{n} \bigg) }^{3} - \dfrac{2n}{n} \bigg]\dfrac{2}{n}[(

n

2

)

3

−

n

2

]

n

2

+−−−+[(

n

2n

)

3

−

n

2n

]

n

2

can be re- written as

= \sf \: \bigg[ {\bigg( \dfrac{2 \times 1}{n} \bigg) }^{3} - \dfrac{2 \times 1}{n} \bigg]\dfrac{2}{n} + \bigg[ {\bigg( \dfrac{2 \times 2}{n} \bigg) }^{3} - \dfrac{2 \times 2}{n} \bigg]\dfrac{2}{n} + - - - + \bigg[ {\bigg( \dfrac{2 \times n}{n} \bigg) }^{3} - \dfrac{2 \times n}{n} \bigg]\dfrac{2}{n}=[(

n

2

Solution−

\large\underline{\sf{Given \: that- }}

Giventhat−

\sf \: \bigg[ {\bigg( \dfrac{2}{n} \bigg) }^{3} - \dfrac{2}{n} \bigg]\dfrac{2}{n} + - - - + \bigg[ {\bigg( \dfrac{2n}{n} \bigg) }^{3} - \dfrac{2n}{n} \bigg]\dfrac{2}{n}[(

n

2

)

3

−

n

2

]

n

2

+−−−+[(

n

2n

)

3

−

n

2n

]

n

2

can be re- written as

= \sf \: \bigg[ {\bigg( \dfrac{2 \times 1}{n} \bigg) }^{3} - \dfrac{2 \times 1}{n} \bigg]\dfrac{2}{n} + \bigg[ {\bigg( \dfrac{2 \times 2}{n} \bigg) }^{3} - \dfrac{2 \times 2}{n} \bigg]\dfrac{2}{n} + - - - + \bigg[ {\bigg( \dfrac{2 \times n}{n} \bigg) }^{3} - \dfrac{2 \times n}{n} \bigg]\dfrac{2}{n}=[(

n

2×1

)

3

−

n

2×1

]

n

2

+[(

n

2×2

)

3

−

n

2×2

]

n

2

+−−−+[(

n

2×n

)

3

−

n

2×n

]

n

2

can be written in summation form as

\sf \: = \displaystyle\:\sum_{i=1}^n \sf \: \bigg[ {\bigg( \dfrac{2i}{n} \bigg) }^{3} - \dfrac{2i}{n} \bigg]\dfrac{2}{n}=

i=1

∑

n

[(

n

2i

)

3

−

n

2i

]

n

2