exponetial & logarithmic series

5/1.2.3 +7/3.4.5 +9/5.6.7 .......... is equal to:
a) log 8/e
b)log e/8
c) log mn
d) none of these

3 Answers

1
Geethu Mg ·

Best Answer: The given series => ( 5 / 1.2.3 ) + (7 / 3.4.5) + ( 9 / 5.6.7) + . . + . . . ∞ Its nth term Tn = (2n + 3) / [ (2n - 1)(2n)(2n + 1) ] Resolving into partial fractions, we get, (2n + 3) / [ (2n - 1)(2n)(2n + 1) ] = 2 / (2n - 1) - 3/ (2n) + 1 / (2n + 1) => Tn = 2 / (2n - 1) - 3/ (2n) + 1 / (2n + 1) n = ∞ . . . . . . . n = ∞ . . ∑ (Tn) = . . ∑ [ 2 / (2n - 1) - 3/ (2n) + 1 / (2n + 1) ] n = 1. . . . . . . .n = 1 n = ∞ . ∑ [ 2 / (2n - 1) - 2 / (2n)] + [1 / (2n + 1) - 1 / 2n ] n = 1 n = ∞ . . . . . . . . . . . . . . . . . . . .n = ∞ 2 * ∑ [ 1 / (2n - 1) - 1 / (2n)] + ∑ [1 / (2n + 1) - 1 / 2n ] n = 1. . . . . . . . . . . . . . . . . . . . n = 1 2 * [ 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + . . . .∞ ] + [ 1/3 - 1/2 + 1/5 - 1/4 + 1/7 - 1/6 + . . . . .∞ ] => 2 * ln 2 + ln 2 - 1 => 3 * ln 2 - 1 ==> (Ans) => 1.079441541. . . . . . . . (Ans)

64
Lokesh Verma ·

\frac{2r+3}{r.(r+1)(r+2)}=\frac{1}{r.(r+1)}+\frac{1}{r.(r+2)}

Hint:

Now try..

1
sindhu br ·

oh!!
thanks

Your Answer

Close [X]