ধারা
প্রমাণ কর যে , 11.2+12.3+13.4…………..=nn+1 \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4} \ldots \ldots \ldots \ldots . .=\frac{n}{n+1} 1.21+2.31+3.41…………..=n+1n
11.2+12.3+13.4…………...=nn+1;Un=1n(n+1)∴Sn=C−1(n+1) \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4} \ldots \ldots \ldots \ldots . . .=\frac{n}{n+1} ; U_{n}=\frac{1}{n(n+1)} \quad \therefore S_{n}=C-\frac{1}{(n+1)} 1.21+2.31+3.41…………...=n+1n;Un=n(n+1)1∴Sn=C−(n+1)1
When, n=0;Sn=0∴0=C−11∴C=1∴Sn=1−1n+1=n+1−1n+1=nn+1 \mathrm{n}=0 ; \mathrm{S}_{\mathrm{n}}=0 \quad \therefore 0=\mathrm{C}-\frac{1}{1} \quad \therefore \mathrm{C}=1 \quad \therefore \mathrm{S}_{\mathrm{n}}=1-\frac{1}{\mathrm{n}+1}=\frac{\mathrm{n}+1-1}{\mathrm{n}+1}=\frac{\mathrm{n}}{\mathrm{n}+1} n=0;Sn=0∴0=C−11∴C=1∴Sn=1−n+11=n+1n+1−1=n+1n
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
