Find the specified nth term in the expansion of the binomial.(x – 10z)7, n = 6Step 1According to the Binomial Theorem,(a + b)n = an + nC1an − 1b + ... + nCran − rbr + ... + nCn − 1abn − 1 + bnthe given expression is already in the form (a + b)n, where a = x, b = $$−10z and n = 6 6 .Step 2Now use the Binomial Theorem to find the 6th term in the expansion of ((x) + (−10z))7. Note that the rth term is nCr − 1an − (r − 1)br − 1 and not nCran − rbr. Also recall that nCr = n!r!(n − r)!, and that (ab)n = an bn.Therefore, the 6th term in the expansion of (x − 10z)7 is as follows.7C5x7 − 5(−10z)5= 7!5! !x2(−10z)5 = 7 · 6 · 5!5! 2!x2(−10)5(z)5 = 7 · 62!x2(−100,000z5) = 7 · 62 · 1x2(−100,000z5) = x2(−100,000z5) =

Now use the Binomial Theorem to find the 6th term in the expansion of ((x) + (−10z))7. Note that the rth term is nCr − 1an − (r − 1)br − 1 and not nCran − rbr. Also recall that nCr = n!r!(n − r)!, and that (ab)n = an bn.Therefore, the 6th term in the expansion of (x − 10z)7 is as follows.7C5x7 − 5(−10z)5= 7!5! !x2(−10z)5 = 7 · 6 · 5!5! 2!x2(−10)5(z)5 = 7 · 62!x2(−100,000z5) = 7 · 62 · 1x2(−100,000z5) = x2(−100,000z5) =

According to the Binomial Theorem,(a + b)n = an + nC1an − 1b + ... + nCran − rbr + ... + nCn − 1abn − 1 + bnthe given expression is already in the form (a + b)n, where a = x, b = and n = .

Write the coefficient of x5in the expansion of: (a) (3 + x)6 (b) (2 − x)7 (c)

If the expansion of  (1 + x)m(1 − x)n , the coefficients of  x  and  x2  are 3 and -6 respectively, then:

In the Binomial expansion of (1+ax)k where a and k are non-zero constants, the coefficient of x is 8 and the coefficient of x2 is 30 a.Determine the value of a and k b.Find the coefficient of x3