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if f(x+y)=f(x)+f(y) and f(1)=7 then ∑f(r) from r=1 to r=n?
Dear student if f(x+y)=f(x)+f(y)Then f(x) = e^(kx) Hence e^k = 7k = ln 7 Hence f(x) = e(ln7)x = 7x Hence from r = 1 to n sigma f(r) = 7 + 7² + 7³ + 7⁴ +.....7ⁿ = 7 ( 7ⁿ - 1) /(7-1) = 7(7ⁿ - 1)/6
Dear studentf(1) = 7f(2) = 2* f(1) = 2*7f(3) = f(2)+f(1) = 2*7 +7...f(n) = n* f(1) = 7*nNow sum can be obtainedRegards
Note that f(r+1)= f(r)+f(1)Or f(r+1)-f(r)= f(1), Which is an AP for r belonging to natural numbers.Hence, f(r)= f(1)+(r-1)*f(1)Or f(r)= r*f(1)= 7rHence summation from 1 to n isS= 7n(n+1)/2
Dear student If f is a function satisfying f(x+y)=f(x).f(y)∀x,y∈N such that f(1)=3and∑x=1nf(x)=120, then find the value of n.3456Sum of n terms of G.P. =Sn=arn−1r−1Given: f(x+y)=f(x).f(y),∑nx=1f(x)=120 and f(1)=3.when x=y=1, f(1+1)=f(2)=f(1).f(1)=3×3=9when x=2andy=1,f(2+1)=f(3)=f(2).f(1)=9×3=27NowGiven: ∑nx=1f(x)==f(1)+f(2)+f(3)+........f(n)=120⇒3+9+27+........f(n)=120This series is a G.P. with a=3,r=3andSn=120We know that Sum of n terms of G.P. =Sn=arn−1r−1⇒3.3n−13−1=120⇒3n−1=40×2=80⇒3n=81=34 ⇒n=4.
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