It is as hard as it doesn't seems!! 2

Algebra Level 4

For a n n degree polynomial f ( x ) f(x) with leading coefficient as 1 \text{leading coefficient as 1}

f ( 1 ) = 0 , f ( 2 ) = 2 , f ( 3 ) = 6 , f ( 4 ) = 12 … pattern continues up to f(n) f(1)=0, f(2)=2, f(3)=6, f(4)=12 \ldots \text{pattern continues up to f(n)}

Find f ( n + 1 ) f(n+1)

a ! = a ( a − 1 ) ( a − 2 ) ( a − 3 ) . . . . 3 × 2 × 1 a!=a(a-1)(a-2)(a-3)....3×2×1

1 n n²+n-n! n! n²+n-(n+1)! n²+n+n! 0 n+1

Aryan Sanghi by Aryan Sanghi , 16, India

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1 solution

Aryan Sanghi
May 17, 2018

Polynomial f ( x ) = ( x − 1 ) ( x − 2 ) ( x − 3 ) . . . . . ( x − n ) + ( x ² − x ) f(x)=(x-1)(x-2)(x-3).....(x-n)+(x²-x)

f ( n + 1 ) = n ( n − 1 ) ( n − 2 ) … 3 × 2 × 1 + ( n + 1 ) ² − ( n + 1 ) f(n+1)=n(n-1)(n-2)\ldots 3×2×1+(n+1)²-(n+1)

f ( n + 1 ) = n ! + n ² + n \boxed{f(n+1)=n!+n²+n}

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