Let a,b,c,d be real numbers such that \sum^{n}_{k=1}(ak^3+bk^2+ck+d)=n^4 for every natural number n. Then |a| + |b| + |c| + |d | is equal to

KVPY SX SB Exam 2013 Question Paper

Question : 82 of 120

Marks: +1, -0

Let a,b,c,d be real numbers such that

∑↖{n}↙{k=1}(ak^3+bk^2+ck+d)=n^4

for every natural number n. Then

|a| + |b| + |c| + |d |

is equal to

Solution:

∑↖{n}↙{k=1}(ak^3+bk^2+ck+d)=n^4

a∑↖{n}↙{k=1}k^3+b∑↖{n}↙{k=1}k^2

+c∑↖{n}↙{k=1}k+∑↖{n}↙{k=1}d=n^2

n^4(12 - 3a) - n^3(4b + 6a)

- n^2(6c + 6b + 3a)

- n(6c + 2b + 12d) = 0

12 - 3a = 0

4b + 6a = 0

6c + 6b + 3a = 0

6c + 2b + 12d = 0

⇒ a = 4, b = - 6

c = 4, d = - 1

|a|+|b|+|c|+|d| = 15

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