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Question : 1 of 160
Marks:
+1,
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Solution:
Given,
f(x+y)=f(x)+f(y) and
f(l)=10=1×10 Then,
f(2)=f(1+1)=f(1)+f(1)=10+10=2×10 f(3)=f(2+1)=f(2)+f(1)=20+10=3×10 f(4)=f(3+1)=f(3)+f(1)=30+10=4×10 ...f(n)=n⋅10 Now,
‌[f(r)]2=[f(1)]2+[f(2)]2+[f(3)]2+...+[f(n)]2 ‌‌=(1⋅10)2+(2⋅10)2+(3⋅10)2+...+(n⋅10)2 ‌‌=102(12+23+32+...+n2) ‌‌=100×‌‌‌[∵∑n2=‌] ‌‌=‌n(n+1)(2n+1)
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