Question

$f(x+1)=(-1{)}^{x+1}x-2f(x)$ for $x\in N$ and $f\left(1\right)=f\left(1986\right)$. Then sum of digits of $\left(f\right(1)+f(2)+....+f(1985\left)\right)=$

• A. $4$
• B. $3$
• C. $7$
• D. $11$

Answer 1

The correct option is C $7$

$f(x+1)-f\left(x\right)=(-1{)}^{x+1}x-3f(x)$

$f\left(2\right)-f\left(1\right)=(-1{)}^2.1-3f(1)$

$f\left(3\right)-f\left(2\right)=(-1{)}^3.2-3f(2)$

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$f\left(1986\right)-f\left(1985\right)=(-1{)}^{1986}x.1985-3f(1985)$

on adding all these we get,

$f\left(1986\right)-f\left(1\right)={\sum }_{n=1}^{1986}{(-1)}^n.n-3{\sum }_{n=1}^{1986}f\left(x\right)$

since $f\left(1986\right)=f\left(1\right)$

$1-2+3-4+5-\mathrm{6.....}+1985-1986=3{\sum }_{n=1}^{1986}f\left(x\right)$

so ${\sum }_{n=1}^{1986}f\left(x\right)=-331$

Thus, sum of digit is $7$.