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Question: Let a, b, c, d be real numbers such that \[\sum\limits_{k=1}^{n}{(a{{k}^{3}}+b{{k}^{2}}+ck+d)}={{n}^...

Let a, b, c, d be real numbers such that ∑k=1n(ak3+bk2+ck+d)=n4\sum\limits_{k=1}^{n}{(a{{k}^{3}}+b{{k}^{2}}+ck+d)}={{n}^{4}} for every natural number n. Then ∣a∣ + ∣b∣ + ∣c∣ + ∣d∣ is equal to:
(a)15
(b)16
(c)31
(d)32

Explanation

Solution

Hint: Use the following formulae to expand the LHS: Sum of the cubes of first n natural numbers = ∑k=1nk3=n2(n+1)24\sum\limits_{k=1}^{n}{{{k}^{3}}}=\dfrac{{{n}^{2}}{{\left( n+1 \right)}^{2}}}{4}, Sum of the squares of first n natural numbers = ∑k=1nk2=n(n+1)(2n+1)6\sum\limits_{k=1}^{n}{{{k}^{2}}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}, Sum of the first n natural numbers = ∑k=1nk=n(n+1)2\sum\limits_{k=1}^{n}{k}=\dfrac{n\left( n+1 \right)}{2}. Then simplify the LHS and compare with the RHS to get the values of a, b, c, d. Then find the final answer.

Complete Step-by-step answer:
We are given that a, b, c, d are real numbers such that ∑k=1n(ak3+bk2+ck+d)=n4\sum\limits_{k=1}^{n}{(a{{k}^{3}}+b{{k}^{2}}+ck+d)}={{n}^{4}} for every natural number n.
We need to find the value of ∣a∣ + ∣b∣ + ∣c∣ + ∣d∣.
We will be using the following formulae:
Sum of the cubes of first n natural numbers = ∑k=1nk3=n2(n+1)24\sum\limits_{k=1}^{n}{{{k}^{3}}}=\dfrac{{{n}^{2}}{{\left( n+1 \right)}^{2}}}{4}
Sum of the squares of first n natural numbers = ∑k=1nk2=n(n+1)(2n+1)6\sum\limits_{k=1}^{n}{{{k}^{2}}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}
Sum of the first n natural numbers = ∑k=1nk=n(n+1)2\sum\limits_{k=1}^{n}{k}=\dfrac{n\left( n+1 \right)}{2}
Substituting these in the equation given in the question, we have the following:
∑k=1n(ak3+bk2+ck+d)=n4\sum\limits_{k=1}^{n}{(a{{k}^{3}}+b{{k}^{2}}+ck+d)}={{n}^{4}}
an2(n+1)24+bn(n+1)(2n+1)6+cn(n+1)2+dn=n4\dfrac{a{{n}^{2}}{{\left( n+1 \right)}^{2}}}{4}+\dfrac{bn\left( n+1 \right)\left( 2n+1 \right)}{6}+\dfrac{cn\left( n+1 \right)}{2}+dn={{n}^{4}}
3an2(n2+2n+1)+2bn(2n2+3n+1)+6cn(n+1)+12dn12=n4\dfrac{3a{{n}^{2}}\left( {{n}^{2}}+2n+1 \right)+2bn\left( 2{{n}^{2}}+3n+1 \right)+6cn\left( n+1 \right)+12dn}{12}={{n}^{4}}
3an2(n2+2n+1)+2bn(2n2+3n+1)+6cn(n+1)+12dn=12n43a{{n}^{2}}\left( {{n}^{2}}+2n+1 \right)+2bn\left( 2{{n}^{2}}+3n+1 \right)+6cn\left( n+1 \right)+12dn=12{{n}^{4}}
3an4+6an3+3an2+4bn3+6bn2+2bn+6cn2+6cn+12dn=12n43a{{n}^{4}}+6a{{n}^{3}}+3a{{n}^{2}}+4b{{n}^{3}}+6b{{n}^{2}}+2bn+6c{{n}^{2}}+6cn+12dn=12{{n}^{4}}
3an4+n3(6a+4b)+n2(3a+6b+6c)+n(2b+6c+12d)=12n43a{{n}^{4}}+{{n}^{3}}\left( 6a+4b \right)+{{n}^{2}}\left( 3a+6b+6c \right)+n\left( 2b+6c+12d \right)=12{{n}^{4}}
On comparing both the sides, we will get the following:
3a = 12 or a = 4 …(1)
6a + 4b = 0
Putting a = 4 in this equation, we get the following:
b = -6 …(2)
3a + 6b + 6c = 0
Putting (1) and (2) in this equation, we get the following:
12 – 36 + 6c = 0 or c = 4 …(3)
2b + 6c + 12d = 0
Putting (2) and (3) in this equation, we will get the following:
-12 + 24 + 12d = 0 or d = 1 …(4)
From (1), (2), (3), and (4),
We have |a| = 4
|b| = 6
|c| = 4
And |d| = 1
So, ∣a∣ + ∣b∣ + ∣c∣ + ∣d∣ = 4 + 6 + 4 + 1 = 15
Hence, ∣a∣ + ∣b∣ + ∣c∣ + ∣d∣ = 15
So, option (a) is correct.

Note: To solve this question, it is very important to know the following formulae: Sum of the cubes of first n natural numbers = ∑k=1nk3=n2(n+1)24\sum\limits_{k=1}^{n}{{{k}^{3}}}=\dfrac{{{n}^{2}}{{\left( n+1 \right)}^{2}}}{4}, Sum of the squares of first n natural numbers = ∑k=1nk2=n(n+1)(2n+1)6\sum\limits_{k=1}^{n}{{{k}^{2}}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}, Sum of the first n natural numbers = ∑k=1nk=n(n+1)2\sum\limits_{k=1}^{n}{k}=\dfrac{n\left( n+1 \right)}{2}.
Just multiplying and simplifying will not get required results hence the above mentioned formulae are very important.