`1+2. 2+3. 2^2+4. 2^3+...+n*2^(n-1)=` (i)`1+(1+n)2^n` (ii)`1-(1+n)2^n` (iii)`1-(1-n)2^n` (iv) `1+(1-n)2^n`

Monosomics are: 1. n 2. 2n +1 3. 2n - 2 4. 2n - 1

1.2.3+2.3.4+....+n(n+1)(n+2)=(n(n+1)(n+2)(n+3))/4

1.2+2.3+3.4+.........+n(n+1)=(1)/(3)n(n+1)(n+3)

Prove that 1*2+2*3+3*4+.....+n*(n+1)=(n(n+1)(n+2))/(3)

Match the following . {:(,"ColumnI",,"ColumnII"),((i) ,1^(2) +2^(2) +3^(2) +....+n^(2) ,(a) ,[(n(n+1))/(2)]^(2)),((ii) , 1^(3) +2^(2) +3^(2) +...+n^(3) ,(b), n(n+1)),((iii),2+4+6+...+2n,( c),(n(n+1)(2n+1))/(6)),((iv),1+2+3+...+n,(d),(n(n+1))/(2)):}

Match the following . {:(,"ColumnI",,"ColumnII"),((i) ,1^(2) +2^(2) +3^(2) +....+n^(2) ,(a) ,[(n(n+1))/(2)]^(2)),((ii) , 1^(3) +2^(3) +3^(3) +...+n^(3) ,(b), n(n+1)),((iii),2+4+6+...+2n,( c),(n(n+1)(2n+1))/(6)),((iv),1+2+3+...+n,(d),(n(n+1))/(2)):}

If A=([x,x],[x,x]) then A^(n)(n in N)= 1) ([2^nx^n,2^nx^n],[2^nx^n,2^nx^n]) 2) ([2^(n-1) x^n,2^(n-1) x^n],[2^(n-1) x^n,2^(n-1) x^n]) 3) I 4) ([2^(n) x^(n-1),2^(n) x^(n-1)],[2^(n) x^(n-1),2^(n) x^(n-1)])

Show : 2^(n)-(n)/(1!).2^(n-1)+(n(n-1))/(2!).2^(n-2)-....+(-1)^(n)=1

1+n. (2n)/(1+n)+(n(n+1))/(1.2) ((2n)/(1+n))^2+…..

1+n. (2n)/(1+n)+(n(n+1))/(1.2) ((2n)/(1+n))^2+…..