Question

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The first term of an arithmetic progression of consecutive integers is k² + 1. The sum of 2k + 1 terms of this progression may be expressed as
(1) k³ + (k + 1)³
(2) (k-1)³+ k³
(3) (K + 1)³
(4) (k + 1)²​

Answer 1

♣ Given :-

For an arithmetic progression of consecutive integers :

• First Term = k² + 1.

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♣ To Find :-

• Sum of 2k + 1 Terms

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♣ Formula for Sum :-

If an A.P contains n terms then such is n terms is given by :

$\large \mathtt{S_n = \dfrac{n}{2} \bigg \{2a + (n - 1)d \bigg \} }$

Where ,

• a = First Term

• d = Common Difference

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♣ Solution :-

Here ,

• a = k² + 1

• d = 1 ( A.P. is consecutive Integers)

• n = 2k + 1

Hence By Formula For Sum ;

$\sf{Sum = \dfrac{2k + 1}{2} } \bigg \{ \sf{2( {k}^{2} + 1) +( 2k + 1 - 1)1 }\bigg \} \\ \\ \large \sf{ = \frac{2k + 1}{2} \bigg(2k {}^{2} + 2 + 2k \bigg) } \\ \\ \large\sf{ = (2k + 1)( {k}^{2} + k + 1)} \\ \\ \large\sf{ =2k^3+2k^2+2k+k^2+k+1 }\\ \\ \large \sf{=2k^3+3k^2+3k+1} \\ \\ \large \sf{ = {k}^{3} + ( {k}^{3} + 1 + 3 {k}^{2} + 3k)} \\ \\ \pink{ \Large \bf = {k}^{3} + ( {k + 1)}^{3} }$

$\underline{ \underline{ \purple {\pmb{\mathcal{Hence \: \: Option \: \: 1 \: \: is \: \: Correct} }}}}$

$\Large \red{\mathfrak{  \text{W}hich \:   \: is  \:  \: the  \:  \: required} }\\ \huge \red{\mathfrak{ \text{ A}nswer.}}$

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Answer 2

Given :-

The first term of an arithmetic progression of consecutive integers is k² +  

To Find :-

The sum of 2k + 1 terms of this progression may be expressed as

Solution :-

We know that

Sₙ = n/2[2a + (n - 1)d]

Sₙ = 2k + 1/2[2(k² + 1) + (2k + 1 - 1)1]

Sₙ = 2k + 1/2 [2k² + 2 + 2k + 1 - 1]

Sₙ = 2k + 1/2[2k² + 2 + 2k]

Sₙ = 2k + 1 × 2k² + 2 + 2k/2

Sₙ = 2k + 1 × k² + 1 + k

Sₙ = 2k³ + 2k² + 2k + k² + k + 1

Sₙ = 2k³ + 3k³ + 3k + 1

Sₙ = k³ + (k + 1)³