ধারা
পাশের ধারাটির n-তম পদ পর্যন্ত যােগফল নির্ণয় কর:
141.4.2.5+684.7.5.8+1587.10.8.11+…….. \frac{14}{1.4 . 2.5} + \frac{68}{4.7 . 5.8} + \frac{158}{7.10 . 8.11} + \ldots \ldots . . 1.4.2.514+4.7.5.868+7.10.8.11158+……..
un=18n2−4(9n2+3n−2)(9n2−3n−2)=(9n2+3n−2)+(9n2−3n−2)(9n2+3n−2)(9n2−3n−2)=1(9n2−3n−2)+1(9n2+3n−2) u_{n}=\frac{18 n^{2}-4}{\left(9 n^{2}+3 n-2\right)\left(9 n^{2}-3 n-2\right)}=\frac{\left(9 n^{2}+3 n-2\right)+\left(9 n^{2}-3 n-2\right)}{\left(9 n^{2}+3 n-2\right)\left(9 n^{2}-3 n-2\right)}=\frac{1}{\left(9 n^{2}-3 n-2\right)}+\frac{1}{\left(9 n^{2}+3 n-2\right)} un=(9n2+3n−2)(9n2−3n−2)18n2−4=(9n2+3n−2)(9n2−3n−2)(9n2+3n−2)+(9n2−3n−2)=(9n2−3n−2)1+(9n2+3n−2)1
=1(3n=1)(3n+2)+1(3n+1)(3n−2)⇒Sn=c−13(3n−1)−13(3n−2)⇒141×4×2×5=c−13.2−13.1∴c=1720∴Sn=1720−13(3n−1)−13(3n−2) \begin{array}{l} =\frac{1}{(3 n=1)(3 n+2)}+\frac{1}{(3 n+1)(3 n-2)} \Rightarrow S_{n}=c-\frac{1}{3(3 n-1)}-\frac{1}{3(3 n-2)} \Rightarrow \frac{14}{1 \times 4 \times 2 \times 5}=c-\frac{1}{3.2}-\frac{1}{3.1} \\ \therefore c=\frac{17}{20} \quad \therefore S_{n}=\frac{17}{20}-\frac{1}{3(3 n-1)}-\frac{1}{3(3 n-2)} \end{array} =(3n=1)(3n+2)1+(3n+1)(3n−2)1⇒Sn=c−3(3n−1)1−3(3n−2)1⇒1×4×2×514=c−3.21−3.11∴c=2017∴Sn=2017−3(3n−1)1−3(3n−2)1
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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 --
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+...
