Maths Experts Please Do it#Help Needed The first term of an arithmetic

1.	Maths Experts Please Do it#Help Needed 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» ♣ Given :- For an arithmetic progression of CONSECUTIVE integers : FIRST Term = k² + 1. ----------------------- ♣ To Find :- Sum of 2k + 1 TERMS ----------------------- ♣ 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 ----------------------- ♣ Solution :- Here , a = k² + 1 d = 1 ( A.P. is consecutive Integers) n = 2k + 1 Hence By Formula For Sum ; $\sf{Sum = \dfrac{2k + 1}{2} } \bigg \{ \sf{2( {k}^{2} + 1) +( 2k + 1 - 1)1 }\bigg \} \\ \\ \large \sf{ = \frac{2k + 1}{2} \bigg(2k {}^{2} + 2 + 2k \bigg) } \\ \\ \large\sf{ = (2k + 1)( {k}^{2} + k + 1)} \\ \\ \large\sf{ =2k^3+2k^2+2k+k^2+k+1 }\\ \\ \large \sf{=2k^3+3k^2+3k+1} \\ \\ \large \sf{ = {k}^{3} + ( {k}^{3} + 1 + 3 {k}^{2} + 3k)} \\ \\ \pink{ \Large \bf = {k}^{3} + ( {k + 1)}^{3} }$ $\underline{ \underline{ \purple {\pmb{\mathcal{Hence \: \: Option \: \: 1 \: \: is \: \: Correct} }}}}$ $\Large \red{\mathfrak{ \text{W}hich \: \: is \: \: the \: \: required} }\\ \huge \red{\mathfrak{ \text{ A}nswer.}}$ -----------------------

Maths Experts Please Do it#Help Needed 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»

♣ Given :-

For an arithmetic progression of CONSECUTIVE integers :

FIRST Term = k² + 1.

-----------------------

♣ To Find :-

Sum of 2k + 1 TERMS

-----------------------

♣ 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

-----------------------

♣ Solution :-

Here ,

a = k² + 1

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

n = 2k + 1

Hence By Formula For Sum ;

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

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

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