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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
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Answers
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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 \} }S
n
=
2
n
{2a+(n−1)d}
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 ;
\begin{gathered}\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} }\end{gathered}
Sum=
2
2k+1
{2(k
2
+1)+(2k+1−1)1}
=
2
2k+1
(2k
2
+2+2k)
=(2k+1)(k
2
+k+1)
=2k
3
+2k
2
+2k+k
2
+k+1
=2k
3
+3k
2
+3k+1
=k
3
+(k
3
+1+3k
2
+3k)
=k
3
+(k+1)
3
\underline{ \underline{ \purple {\pmb{\mathcal{Hence \: \: Option \: \: 1 \: \: is \: \: Correct} }}}}
HenceOption1isCorrect
HenceOption1isCorrect
\begin{gathered}\Large \red{\mathfrak{ \text{W}hich \: \: is \: \: the \: \: required} }\\ \huge \red{\mathfrak{ \text{ A}nswer.}}\end{gathered}
Which is the required
Answer.
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