Dfferentiate the maclaurin series 1(1−6x) twice to find the maclaurin series of 1(1−6x)3. Index the series so that the 0th term is nonzero

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jlo2niT · Accepted Answer

The objective here is to find the maclaurin series of1(1−6x)3Twice differentiating1(1−6x)gives:[1(1−6x)]′=−6(1−6x)2[1(1−6x)]″=[−6(1−6x)2]′=−6(6)(−2)(1−6x)3=72(1−6x)3Also note that the11−r is represented bythe infinite geometric series sum11−r=∑n=0∞rn⇒11−6x=∑n=0∞(6x)nNow, twice differentiating ∑n=0∞(6x)n(∑n=0∞(6x)n)′=∑n=1∞6nn(x)n−1 and(∑n=0∞(6x)n)″=(∑n=1∞6nn(x)n−1)′=∑n=2∞6nn(n−1)(x)n−2(∑n=0∞(6x)n)″=∑n=2∞6nn(n−1)(x)n−2And since[1(1−6x)]″=72(1−6x)3Thus, 72(1−6x)3=∑n=2∞6nn(n−1)(x)n−2Since, the above series starts with n=2 so re-index this so that it starts with 0 by adding initial two terms72(1−6x)3=∑n=2∞6nn(n−1)(x)n−2=∑n=0∞(n+2)(n+2−1)6n+2(x)n−2+2=∑n=0∞(n+1)(n+2)6n+2(x)nThus,1(1−6x)3=172∑n=0∞(n+1)(n+2)6n+2(x)n=∑n=0∞(n+1)(n+2)6n+2(x)n2⋅62=∑n=0∞(n+1)(n+2)6n(x)n2Thus, required Maclaurin series expansion is1(1−6x)3=∑n=0∞(n+1)(n+2)6n(x)n2Note that here 0th term is nonzero

Jeffrey Jordon · Accepted Answer

Answer is given below (on video)

Dfferentiate the maclaurin series frac{1}{(1-6x)} twice to find the maclaurin series of frac{1}{(1-6x)^3}. Index the series so that the 0th term is nonzero

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