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Question: The first term of an A.P of consecutive integers is \[{p^2} + 1\] . The sum of \[(2p + 1)\]terms of ...

The first term of an A.P of consecutive integers is p2+1{p^2} + 1 . The sum of (2p+1)(2p + 1)terms of this series can be expressed as
A. (p+1)2{(p + 1)^2}
B. (p+1)3{(p + 1)^3}
C. (2p+1)(p+1)2(2p + 1){(p + 1)^2}
D. p3+(p+1)3{p^3} + {(p + 1)^3}

Explanation

Solution

since first term of the arithmetic progression of consecutive integers and the sum of terms is given so we can solve the given problem by using the formula for sum of first n term of arithmetic progression.by taking n as 2p+12p + 1 as sum of 2p+12p + 1terms is given in the problem.

Complete step by step answer:
Given: The first term of three consecutive integers of A.P=p2+1 = {p^2} + 1
Let the three consecutive integers be of the form (p2+1),(p2+2),(p2+3).({p^2} + 1),({p^2} + 2),({p^2} + 3).
Now, sum of n terms of A.P is given by sn=n2(2a+(n−1)d){s_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)
Where,
a→a \to first term of the A.P
d→d \to common difference
n→n \to number of terms in the given series
Now from the given data
a=p2+1a = {p^2} + 1
n=2p+1n = 2p + 1
d=1d = 1 since integers are consecutive
On substituting the values in the formula, we get
sn=(2p+12)(2(p2+1)+((2p+1)−1)(1)){s_n} = \left( {\dfrac{{2p + 1}}{2}} \right)\left( {2\left( {{p^2} + 1} \right) + \left( {(2p + 1) - 1} \right)(1)} \right)
On simplification we get
sn=(2p+12)(2p2+2p+2){s_n} = \left( {\dfrac{{2p + 1}}{2}} \right)\left( {2{p^2} + 2p + 2} \right)
Now by taking 2 as a common factor the above equation becomes
sn=(2p+12)2(p2+p+1){s_n} = \left( {\dfrac{{2p + 1}}{2}} \right)2\left( {{p^2} + p + 1} \right)
On simplification the above equation can be written as
sn=(2p+1)(p2+p+1){s_n} = (2p + 1)\left( {{p^2} + p + 1} \right)
Now multiplying the terms, we get
sn=2p3+2p2+2p+P2+p+1{s_n} = 2{p^3} + 2{p^2} + 2p + {P^2} + p + 1
Now we can rearrange the terms as per the options given, we can write the above equation as follows
sn=p3+(p3+3p2+3p+1){s_n} = {p^3} + ({p^3} + 3{p^2} + 3p + 1)
Now the terms in the bracket is of the form (a+b)3{(a + b)^3}so we can write the terms in the bracket in the form of (p+1)3{(p + 1)^3}therefore, the above equation can be written as
sn=p3+(p+1)3{s_n} = {p^3} + {(p + 1)^3}

So, the correct answer is “Option D”.

Note: Arithmetic Progression is the most commonly used sequence in mathematics and is defined as a sequence of numbers in order in which the difference of any two consecutive numbers is a constant value. We can also use the formula for sum of the n terms of A.P as sn=n2(firstterm+lastterm){s_n} = \dfrac{n}{2}\left( {first term + last term} \right)if the last term is given.