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Question

# Prove that:(i) 13+√7+1√7+√5+1√5+√3+1√3+1=1(ii) 11+√2+1√2+√3+1√3+√4+1√4+√5+1√5+√6+1√6+√7+1√7+√8+1√8+√9=2

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Solution

## (i)To prove: 13+√7+1√7+√5+1√5+√3+1√3+1=1Lets take LHS and then equate it to RHS.LHS =13+√7+1√7+√5+1√5+√3+1√3+1Lets do rationalize the denominator of each term,=[13+√7×3−√73−√7]+[1√7+√5×√7−√5√7−√5]+[1√5+√3×√5−√3√5−√3]+[1√3+1×√3−1√3−1]=3−√732−(√7)2+√7−√5(√7)2−(√5)2+√5−√3(√5)2−(√3)2+√3−1(√3)2−12=3−√79−7+√7−√57−5+√5−√35−3+√3−13−1=3−√72+√7−√52+√5−√32+√3−12=3−√7+√7−√5+√5−√3+√3−12=22=1= RHS∴ LHS = RHSHence, proved.(ii) To prove:11+√2+1√2+√3+1√3+√4+1√4+√5+1√5+√6+1√6+√7+1√7+√8+1√8+√9=2Lets take LHS and then equate it to RHS.LHS =11+√2+1√2+√3+1√3+√4+1√4+√5+1√5+√6+1√6+√7+1√7+√8+1√8+√9Lets do rationalize the denominator of each term,=[11+√2×√2−1√2−1]+[1√2+√3×√3−√2√3−√2]+[1√4+√3×√4−√3√4−√3]+[1√4+√5×√5−√4√5−√4]+[1√5+√6×√6−√5√6−√5]+[1√6+√7×√7−√6√7−√6]+[1√7+√8×√8−√7√8−√7]+[1√8+√9×√9−√8√9−√8]=√2−1(√2)2−12+√3−√2(√3)2−(√2)2+√4−√3(√4)2−(√3)2+√5−√4(√5)2−(√4)2+√6−√5(√6)2−(√5)2+√7−√6(√7)2−(√6)2+√8−√7(√8)2−(√7)2+√9−√8(√9)2−(√8)2=√2−12−1+√3−√23−2+√4−√34−3+√5−√45−4+√6−√56−5+√7−√67−6+√8−√78−7+√9−√89−8=√2−11+√3−√21+√4−√31+√5−√41+√6−√51+√7−√61+√8−√71+√9−√81=√2−1+√3−√2+√4−√3+√5−√4+√6−√5+√7−√6+√8−√7+√9−√8=√9−1=3−1=2∴ LHS = RHSSo, 11+√2+1√2+√3+1√3+√4+1√4+√5+1√5+√6+1√6+√7+1√7+√8+1√8+√9=2Hence, proved.

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