ধারা
21!+2+42!+2+4+63!+…∞\frac{2}{1!}+\frac{2+4}{2!}+\frac{2+4+6}{3!}+\ldots\infty1!2+2!2+4+3!2+4+6+…∞ ধারাটির যোগফল কত?
3e
4e
2e
5e
21!+2+42!+2+4+63!+…∞ Un=2+4+6+…+2nn!=n(n+1)n!=n+1(n−1)!=(n−1)+2(n−1)!=n−1(n−1)!+2(n−1)!\frac{2}{1!}+\frac{2+4}{2!}+\frac{2+4+6}{3!}+\ldots\infty\ \\ U_n=\frac{2+4+6+\ldots+2n}{n!}=\frac{n\left(n+1\right)}{n!}=\frac{n+1}{\left(n-1\right)!}=\frac{\left(n-1\right)+2}{\left(n-1\right)!}=\frac{n-1}{\left(n-1\right)!}+\frac{2}{\left(n-1\right)!}1!2+2!2+4+3!2+4+6+…∞ Un=n!2+4+6+…+2n=n!n(n+1)=(n−1)!n+1=(n−1)!(n−1)+2=(n−1)!n−1+(n−1)!2
∴Sn=∑n=1n=∞Un=∑n=1n=∞n−1(n−1)!+2∑n=1n=∞1(n−1)!=e+2e=3e \therefore S_n=\overset{n=\infty}{\underset{n=1} \sum}U_n =\overset{n=\infty}{\underset{n=1} \sum} \frac{n-1}{\left(n-1\right)!}+2 \overset{n=\infty}{\underset{n=1} \sum}\frac{1}{\left(n-1\right)!}=e+2e=3e\ ∴Sn=n=1∑n=∞Un=n=1∑n=∞(n−1)!n−1+2n=1∑n=∞(n−1)!1=e+2e=3e
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
