ধারা
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]
(2mn+1)(2n+1)(2mn)\displaystyle\cfrac { \left( { 2 }^{ mn }+1 \right) }{ \left( { 2 }^{ n }+1 \right) \left( { 2 }^{ mn } \right) } (2n+1)(2mn)(2mn+1)
(2mn−1)(2n−1)(2mn)\displaystyle\cfrac { \left( { 2 }^{ mn }-1 \right) }{ \left( { 2 }^{ n }-1 \right) \left( { 2 }^{ mn } \right) } (2n−1)(2mn)(2mn−1)
(2mn+1)(2n−1)(2mn)\displaystyle\cfrac { \left( { 2 }^{ mn }+1 \right) }{ \left( { 2 }^{ n }-1 \right) \left( { 2 }^{ mn } \right) } (2n−1)(2mn)(2mn+1)
(2mn−1)(2n+1)(2mn)\displaystyle\cfrac { \left( { 2 }^{ mn }-1 \right) }{ \left( { 2 }^{ n }+1 \right) \left( { 2 }^{ mn } \right) } (2n+1)(2mn)(2mn−1)
∑r=0n(−1)n.nCr[12r+3r22r+7r23r+15r24r+...]\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 } } +... \right] r=0∑n(−1)n.nCr[2r1+22r3r+23r7r+24r15r+...] upto m terms
=∑r=0n(−1)nnCr(12)r+∑r=0n(−1)nnCr(34)r+∑r=0n(−1)nnCr()(78)r+...\displaystyle=\sum _{ r=0 }^{ n }{ { \left( -1 \right) }^{ n } } { _{ }^{ n }{ C } }_{ r }{ \left( \frac { 1 }{ 2 } \right) }^{ r }+\sum _{ r=0 }^{ n }{ { \left( -1 \right) }^{ n } } { _{ }^{ n }{ C } }_{ r }{ \left( \frac { 3 }{ 4 } \right) }^{ r }+\sum _{ r=0 }^{ n }{ { \left( -1 \right) }^{ n } } { _{ }^{ n }{ C } }_{ r }\left( \right) { \left( \frac { 7 }{ 8 } \right) }^{ r }+...=r=0∑n(−1)nnCr(21)r+r=0∑n(−1)nnCr(43)r+r=0∑n(−1)nnCr()(87)r+...
=(1−12)n+(1−34)n+(1−78)n+...[∵∑r=0n(−1)nnCrxr=(1−x)r]\displaystyle={ \left( 1-\frac { 1 }{ 2 } \right) }^{ n }+{ \left( 1-\frac { 3 }{ 4 } \right) }^{ n }+{ \left( 1-\frac { 7 }{ 8 } \right) }^{ n }+...\quad \quad \left[ \because \sum _{ r=0 }^{ n }{ { \left( -1 \right) }^{ n } } { _{ }^{ n }{ C } }_{ r }{ x }^{ r }={ \left( 1-x \right) }^{ r } \right] =(1−21)n+(1−43)n+(1−87)n+...[∵r=0∑n(−1)nnCrxr=(1−x)r]
=(12)n+(14)n+(18)n+...=(12)n[1−(12n)m1−(12)n]\displaystyle={ \left( \frac { 1 }{ 2 } \right) }^{ n }+{ \left( \frac { 1 }{ 4 } \right) }^{ n }+{ \left( \frac { 1 }{ 8 } \right) }^{ n }+...={ \left( \frac { 1 }{ 2 } \right) }^{ n }\left[ \frac { 1-{ \left( \frac { 1 }{ { 2 }^{ n } } \right) }^{ m } }{ 1-{ \left( \frac { 1 }{ 2 } \right) }^{ n } } \right] =(21)n+(41)n+(81)n+...=(21)n[1−(21)n1−(2n1)m]
Ai এর মাধ্যমে
১০ লক্ষ+ প্রশ্ন ডাটাবেজ
প্র্যাকটিস এর মাধ্যমে নিজেকে তৈরি করে ফেলো
উত্তর দিবে তোমার বই থেকে ও তোমার মত করে।
সারা দেশের শিক্ষার্থীদের মধ্যে নিজের অবস্থান যাচাই
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
Find the value of 1(n−1)!+1(n−3)!3!+1(n−5)!5!+...\dfrac{1}{\left(n-1\right)!}+\dfrac{1}{\left(n-3\right)!3!}+\dfrac{1}{\left(n-5\right)!5!}+...(n−1)!1+(n−3)!3!1+(n−5)!5!1+...
If Tr=2016Crx2016−rT_r=^{2016}C_rx^{2016-r}Tr=2016Crx2016−r, for r=0,1,,....2016r=0, 1, ,....2016r=0,1,,....2016, then (T0−T2+T4....+T2016)2+(T1−T3+T5....T2015)2(T_0 - T_2+T_4....+T_{2016})^2+(T_1-T_3+T_5....T_{2015})^2(T0−T2+T4....+T2016)2+(T1−T3+T5....T2015)2 is equal to - ;
(1+x)15=a0+a1x+……+a15x15⇒∑r=115rarar−1= (1+x)^{15}=a_0+a_1x+\ldots\ldots+a_{15}x^{15} \Rightarrow \sum_{r=1}^{15}r\frac{a_r}{a_{r-1}}= (1+x)15=a0+a1x+……+a15x15⇒∑r=115rar−1ar=