For k- let 1- - -1-2 - -20-k-020Ak- -k- where - -0- Then the value of 100 A14-A15A13 2 is equal to

Α (α +1)(α +2)…(α +20)∑_k=0^20A_kα +k=1 Put α = 14, A_14· 14! 6!=1⇒ A_14=1 6! 14! Put α = 13, A_13· 13! 7!=1⇒ A_13= 1 7! 13! Put α = 15, A_15· 15! 5!= ...

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Standard XII

Mathematics

Infinite GP

For k∈ℕ, let ...

Question

For $k \in N,$ let $\frac{1}{α (α + 1) (α + 2) \dots (α + 20)} = 20 \sum k = 0 \frac{A_{k}}{α + k},$ where $α > 0.$ Then the value of $100 {(\frac{A_{14} + A_{15}}{A_{13}})}^{2}$ is equal to

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Solution

$α (α + 1) (α + 2) \dots (α + 20) 20 \sum k = 0 \frac{A_{k}}{α + k} = 1$
Put $α = - 14, A_{14} \cdot 14! 6! = 1 \Rightarrow A_{14} = \frac{1}{6! 14!}$
Put $α = - 13, - A_{13} \cdot 13! 7! = 1 \Rightarrow A_{13} = \frac{- 1}{7! 13!}$
Put $α = - 15, - A_{15} \cdot 15! 5! = 1 \Rightarrow A_{15} = \frac{- 1}{15! 5!}$

$100 {(\frac{A_{14} + A_{15}}{A_{13}})}^{2} = \frac{100 {(\frac{1}{14! 6!} - \frac{1}{15! 5!})}^{2}}{{(\frac{1}{13! 7!})}^{2}} = 100 {⎛ ⎜ ⎜ ⎜ ⎝ \frac{\frac{9}{15! 6!}}{\frac{1}{7! 13!}} ⎞ ⎟ ⎟ ⎟ ⎠}^{2} = 9$

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For k∈N, let 1α(α+1)(α+2)…(α+20)=20∑k=0Akα+k, where α>0. Then the value of 100(A14+A15A13)2 is equal to

α(α+1)(α+2)…(α+20)20∑k=0Akα+k=1 Put α=−14, A14⋅14! 6!=1⇒A14=16! 14! Put α=−13, −A13⋅13! 7!=1⇒A13=−17! 13! Put α=−15, −A15⋅15! 5!=1⇒A15=−115! 5! 100(A14+A15A13)2=100(114! 6!−115! 5!)2(113! 7!)2=100⎛⎜ ⎜ ⎜⎝915! 6!17! 13!⎞⎟ ⎟ ⎟⎠2=9

For $k \in N,$ let $\frac{1}{α (α + 1) (α + 2) \dots (α + 20)} = 20 \sum k = 0 \frac{A_{k}}{α + k},$ where $α > 0.$ Then the value of $100 {(\frac{A_{14} + A_{15}}{A_{13}})}^{2}$ is equal to