66Hint: Consider the series expansion of the general exponential
a^x = 1 + x\ln a + \dfrac{(x\ln a)^2}{2!}+\dfrac{(x\ln a)^3}{3!}+\ldots
FInd the sum of inifnite series
a1+a2+a3+..... where a_n=(log3)^n\sum_{k=1}^{n} \frac {2k+1}{k!(n-k)!}
ALso it is known that a0,a1,a2.... be the coefficients in the expasion of (1+x+x2)n in ascending power of x
i have calculated a_n=\frac {(2log3)^n}{(n-1)!}+\frac {(2log3)^n}{n!}-\frac {(log3)^n}{n!}
but cant progress further
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6 Answers
66
62Eureka.. The hint is staring in your face :D
Look at each term of what you have given...
Each of them are very similar to 32 ??
See the expansion of 32 using the exapnsino that kaymant sir has given...
106as given by eure.. (not verified)
a_{n} = \frac{2log3(2log3)^n^-^1}{(n-1)!} + \frac{(2log3)^n}{n!} - \frac{(log3)^n}{n!}
summing this.
S = 2log(3).e2log3 + e2log3 - 1 -(elog3-1)
= 18log3 + 9-3
= 18log3 + 6