Find X 1 6 X 1 6 Hence Evaluate 2 1 6 2 1 6

Find X 1 6 X 1 6 Hence Evaluate 2 1 6 2 1 6 - Solution. Verified by Toppr. Express the given number as the sum or difference of two numbers whose powers are easier to evaluate. Then use binomial theorem as follows: (a+b)n =∑n k=0 n kCan−kbk =n 0 Canb0 +n 1 Can−1b1+......n nCa0bn. , where b0 =1 = an−n. Use the above to evaluate, (x+1)6 +(x −1)6. (x+1)6 = 6 0C x610 +6 1 C x511+.... Using Binomial Theorem the expressions x 1 6 and x 1 6 can be expanded as x 1 6 x 6 6C 1 x 5 1 1 6C 2 x 4 xx 1 2 6C 3 x 3 xx 1 3 6C 4 x 2 1 4 6C 5

Find X 1 6 X 1 6 Hence Evaluate 2 1 6 2 1 6

Find X 1 6 X 1 6 Hence Evaluate 2 1 6 2 1 6

Step 1: Enter the expression you want to evaluate. The Math Calculator will evaluate your problem down to a final solution. You can also add, subtraction, multiply, and divide and complete any arithmetic you need. Step 2: Click the. Find (x + 1)⁶ + (x - 1)⁶. Hence or otherwise evaluate (√2 + 1)⁶ + (√2 - 1)⁶. Solution: Using binomial theorem, the expressions (x + 1)⁶ and (x - 1)⁶ can be expanded as (x + 1)⁶ = ⁶C₀ x⁶ + ⁶C₁ x⁵ + ⁶C₂ x⁴ + ⁶C₃ x³ + ⁶C₄ x² + ⁶C₅ x + ⁶C₆

Find x 1 6 x 1 6 Hence Or Otherwise Evaluate 2 1 6 2 1 6

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Find X 1 6 X 1 6 Hence Evaluate 2 1 6 2 1 6Question. Find (x+1) 6+(x−1) 6 . Hence or otherwise evaluate ( 2+1) 6+( 2−1) 6. Medium. Solution. Verified by Toppr. Express the given number as the sum or difference of two numbers whose powers are easier to evaluate. Then use binomial theorem as follows: (a+b) n=∑ k=0n knCa n−kb k= 0nCa nb 0+ 1nCa n−1b 1+...... nnCa 0b n. , where b 0=1=a n−n. Transcript Ex 7 1 12 Find x 1 6 x 1 6 Hence or otherwise evaluate 2 1 6 2 1 6 We know that a b n nC0 an nC1 an 1 b1 nC2 an 2 b2 nCn 1 a1 bn 1 nCn bn Hence a b 6 6 0 6 0 a6 215 1 6 1 6 1 a5 b 6 2 6 2 a4 b2 6 3 6 3 a3 b3 6

Online math solver with free step by step solutions to algebra, calculus, and other math problems. Get help on the web or with our math app. Evaluate 2 1 1 1 Brainly in If X 2 root 3 Then Solve X 3 1 x 3 Brainly in

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