If $$11.2.3.4+12.3.4.5+13.4.5.6+$$…… upto n terms $$=118$$−$$13f(n)$$ then $$f(1)$$−$$f(3)$$ is

The Genereal term of the given series is $$tn=1n(n+1)(n+2)(n+3)$$ Suppose that $$Vr=1r(r+1)(r+2)Vr-Vr+1=3tr$$ Hence, $$tr=13Vr-Vr+1$$∑$$r=1ntn=13V1-Vn+1$$ $$=1316-1n+1n+2n+3$$ $$=118-13fn$$ $$fn=n+1n+2n+3f1-f3=24-120=-96$$

Special Series in Sequences and Series

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$If \frac{1}{1.2.3.4} + \frac{1}{2.3.4.5} + \frac{1}{3.4.5.6} + \dots \dots upto n terms = \frac{1}{18} - \frac{1}{3 f (n)} then f (1) - f (3) is$

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Solution

$T h e G e n e r e a l t e r m o f t h e g i v e n s e r i e s i s t_{n} = \frac{1}{n (n + 1) (n + 2) (n + 3)} S u p p o s e t h a t V_{r} = \frac{1}{r (r + 1) (r + 2)}$
$V_{r} - V_{r + 1} = 3 t_{r} H e n c e, t_{r} = \frac{1}{3} (V_{r} - V_{r + 1})$
$\sum_{r = 1}^{n} t_{n} = \frac{1}{3} (V_{1} - V_{n + 1}) = \frac{1}{3} (\frac{1}{6} - \frac{1}{(n + 1) (n + 2) (n + 3)}) = \frac{1}{18} - \frac{1}{3 f (n)} f (n) = (n + 1) (n + 2) (n + 3)$
$f (1) - f (3) = 24 - 120 = - 96$

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Special Series in Sequences and Series

Remember concepts with our Masterclasses.

If 11.2.3.4+12.3.4.5+13.4.5.6+…… upto n terms =118−13f(n) then f(1)−f(3) is

The Genereal term of the given series is tn=1n(n+1)(n+2)(n+3) Suppose that Vr=1r(r+1)(r+2)Vr-Vr+1=3tr Hence, tr=13Vr-Vr+1∑r=1ntn=13V1-Vn+1 =1316-1n+1n+2n+3 =118-13fn fn=n+1n+2n+3f1-f3=24-120=-96

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$If \frac{1}{1.2.3.4} + \frac{1}{2.3.4.5} + \frac{1}{3.4.5.6} + \dots \dots upto n terms = \frac{1}{18} - \frac{1}{3 f (n)} then f (1) - f (3) is$