If f:R→R such that fx+y=fx+fy for all x,y∈R. f1=7. and ∑r=1nfr=14112 , then n=

fx+y=fx+fyf1=7,f2=14,...fn=7n∑r=1nfr=14112f1+f2+....+fn=1411271+2+3+....+n=141127nn+12=14112⇒nn+1=63×64n=63

If f:R→R such that fx+y=fx+fy for all x,y∈R. f1=7. and ∑r=1nfr=14112 , then n=

fx+y=fx+fyf1=7,f2=14,...fn=7n∑r=1nfr=14112f1+f2+....+fn=1411271+2+3+....+n=141127nn+12=14112⇒nn+1=63×64n=63

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If $f : R \to R$ such that $f (x + y) = f (x) + f (y)$ for all $x, y \in R$ . $f (1) = 7$ . and $\sum_{r = 1}^{n} f (r) = 14112$ , then $n =$

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Detailed Solution

$f (x + y) = f (x) + f (y)$

$f (1) = 7, f (2) = 14, ... f (n) = 7 n$

$\sum_{r = 1}^{n} f (r) = 14112$

$f (1) + f (2) + .... + f (n) = 14112$

$7 (1 + 2 + 3 + .... + n) = 14112$

$7 \frac{n (n + 1)}{2} = 14112 \Rightarrow n (n + 1) = 63 \times 64$

$n = 63$

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If f:R→R such that fx+y=fx+fy for all x,y∈R. f1=7. and ∑r=1nfr=14112 , then n=

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If $f : R \to R$ such that $f (x + y) = f (x) + f (y)$ for all $x, y \in R$ . $f (1) = 7$ . and $\sum_{r = 1}^{n} f (r) = 14112$ , then $n =$