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Let $\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}$ be real numbers such that
$$
\sum_{k=1}^{n}\left(a k^{3}+b k^{2}+c k+d\right)=n^{4}
$$
for every natural number $\mathrm{n}$. Then $|\mathrm{a}|+|\mathrm{b}|+|\mathrm{C}|+|\mathrm{d}|$ is equal to
Options:
$$
\sum_{k=1}^{n}\left(a k^{3}+b k^{2}+c k+d\right)=n^{4}
$$
for every natural number $\mathrm{n}$. Then $|\mathrm{a}|+|\mathrm{b}|+|\mathrm{C}|+|\mathrm{d}|$ is equal to
Solution:
2117 Upvotes
Verified Answer
The correct answer is:
15
$\sum_{k=1}^{n}\left(a k^{3}+b k^{2}+c k+d\right)=n^{4}$
$a \sum_{k=1}^{n} k^{3}+b \sum_{k=1}^{n} k^{2}+c \sum_{k=1}^{n} k+\sum_{k=1}^{n} d=n^{2}$
$n^{4}(12-3 a)-n^{3}(4 b+6 a)-n^{2}(6 c+6 b+3 a)-n(6 c+2 b+12 d)=0$
$12-3 a=0, \quad 4 b+6 a=0, \quad 6 c+6 b+3 a=0, \quad 6 c+2 b+12 d=0$
$\Rightarrow a=4, b=-6, c=4, d=-1$
$|a|+|b|+|c|+|d|=15$
$a \sum_{k=1}^{n} k^{3}+b \sum_{k=1}^{n} k^{2}+c \sum_{k=1}^{n} k+\sum_{k=1}^{n} d=n^{2}$
$n^{4}(12-3 a)-n^{3}(4 b+6 a)-n^{2}(6 c+6 b+3 a)-n(6 c+2 b+12 d)=0$
$12-3 a=0, \quad 4 b+6 a=0, \quad 6 c+6 b+3 a=0, \quad 6 c+2 b+12 d=0$
$\Rightarrow a=4, b=-6, c=4, d=-1$
$|a|+|b|+|c|+|d|=15$
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