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Evaluate the Square Root of (11⁴+100⁴+111⁴)/2 | IOQM

JEE Maths question with a full step-by-step solution.

Question
Calculate the value of 114+1004+11142\sqrt{\dfrac{11^4 + 100^4 + 111^4}{2}}.
Solution
Answer: 11221
Step 1: Key identity. For all real a,ba, b,
a4+b4+(a+b)4=12[a2+b2+(a+b)2]2.a^4 + b^4 + (a + b)^4 = \frac{1}{2}\big[a^2 + b^2 + (a + b)^2\big]^2.
 Step 2: Since 111=11+100111 = 11 + 100, take a=11a = 11, b=100b = 100, a+b=111a + b = 111. Then
114+1004+1114=12[112+1002+1112]2,11^4 + 100^4 + 111^4 = \frac{1}{2}\big[11^2 + 100^2 + 111^2\big]^2,
 so
114+1004+11142=14[112+1002+1112]2.\frac{11^4 + 100^4 + 111^4}{2} = \frac{1}{4}\big[11^2 + 100^2 + 111^2\big]^2.
 Step 3: Take the square root.
114+1004+11142=112+1002+11122=121+10000+123212=224422=11221.\sqrt{\frac{11^4 + 100^4 + 111^4}{2}} = \frac{11^2 + 100^2 + 111^2}{2} = \frac{121 + 10000 + 12321}{2} = \frac{22442}{2} = 11221.
 Answer: 1122111221.
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