ধারা
1+222!+243!+264!+…… 1 + \frac{2^{2}}{2 !} + \frac{2^{4}}{3 !} + \frac{2^{6}}{4 !} + … … 1+2!22+3!24+4!26+…… ধারাটির যোগফল কত?
e−3+14 \frac{e^{- 3} + 1}{4} 4e−3+1
e−4−14 \frac{e^{- 4} - 1}{4} 4e−4−1
e4−14 \frac{e^{4} - 1}{4} 4e4−1
S=1+222!+243!+264!+……………⇒S=1+42!+423!+434!+…………∞ \mathrm{S}=1+\frac{2^{2}}{2 !}+\frac{2^{4}}{3 !}+\frac{2^{6}}{4 !}+\ldots \ldots \ldots \ldots \ldots \Rightarrow \mathrm{S}=1+\frac{4}{2 !}+\frac{4^{2}}{3 !}+\frac{4^{3}}{4 !}+\ldots \ldots \ldots \ldots \infty S=1+2!22+3!24+4!26+……………⇒S=1+2!4+3!42+4!43+…………∞
⇒4 S=4+422!+433!+444!+…………...∞ \Rightarrow 4 \mathrm{~S}=4+\frac{4^{2}}{2 !}+\frac{4^{3}}{3 !}+\frac{4^{4}}{4 !}+\ldots \ldots \ldots \ldots . . . \infty \quad ⇒4 S=4+2!42+3!43+4!44+…………...∞ [উভয় পক্ষ 4 দিয়ে গুণ ।]
⇒4 S+1=1+41!+422!+433!+444!+…………...∞⇒4 S+1=e4⇒S=e4−14 \Rightarrow 4 \mathrm{~S}+1=1+\frac{4}{1 !}+\frac{4^{2}}{2 !}+\frac{4^{3}}{3 !}+\frac{4^{4}}{4 !}+\ldots \ldots \ldots \ldots . . . \infty \Rightarrow 4 \mathrm{~S}+1=\mathrm{e}^{4} \Rightarrow \mathrm{S}=\frac{\mathrm{e}^{4}-1}{4} ⇒4 S+1=1+1!4+2!42+3!43+4!44+…………...∞⇒4 S+1=e4⇒S=4e4−1
Shortcut: Calculator দিয়ে (১২ তম পদ পর্যন্ত) হিসেবে করলে 13.395... আলে যা D এর মানের কাছাকাছি।
Find the sum of the series ∑r=0n(−1)nnCr[12r+3r22r+7r23r+15r24r+...upto m terms]\displaystyle\sum _{ r=0 }^{ n }{ { \left( -1 \right) }^{ n } } { _{ }^{ n }{ C } }_{ r }\left[ \cfrac { 1 }{ { 2 }^{ r } } +\cfrac { { 3 }^{ r } }{ { 2 }^{ 2r } } +\cfrac { { 7 }^{ r } }{ { 2 }^{ 3r } } +\cfrac { { 15 }^{ r } }{ { 2 }^{ 4r } } +...upto\: m\: terms \right] r=0∑n(−1)nnCr[2r1+22r3r+23r7r+24r15r+...uptomterms]
Let n be a positive integer. Then the value of ∑k=0n(−1)k(nk) \sum_{k = 0}^{n} \left ( - 1 \right )^{k} \left ( \frac{n}{k} \right ) ∑k=0n(−1)k(kn) is --
The arithmetic mean of nC0, nC1, nC2...., nCn^nC_0 , \ ^nC_1, \ ^nC_2 ...., \ ^nC_nnC0, nC1, nC2...., nCn is ;
Number of different terms in the sum (1+x)2009⋅(1+x2)2008+(1+x3)2007, ( 1 + x ) ^ { 2009 } \cdot \left( 1 + x ^ { 2 } \right) ^ { 2008 } + \left( 1 + x ^ { 3 } \right) ^ { 2007 } , (1+x)2009⋅(1+x2)2008+(1+x3)2007, is