∵f(x+y)=f(x)+f(y)
∴f(x)=kx,∀x∈R
Now f(1)=2⇒k=2
∴f(x)=2x,∀x∈R
g(n)=f(1)+f(2)+……..+f(n−1)
=2[1+2+3+……+(n−1)]
=2⋅2(n−1)n=20
⇒n2−n−20=0
⇒n=5.
JEE Main 2020 — Mathematics Algebra
Let f:R→R be a function which satisfies f(x+y)=f(x)+f(y),∀x,y∈R . If f(1)=2 and g(n)=k=1∑(n−1)f(k),n∈N then the value of n, for which g(n)=20, is
Held on 2 Sept 2020 · Verified 6 Jul 2026.
5
20
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9
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