Question

find 2×6+4×9+6×12+.........upto n terms ​

Answer 1

Answer:

2×6+ 4×9 + 6× 12+..... n terms

This can bedoneby applying summation by (2+4+...) and .(6+9+...) as 2 different APS

but let's do it in unique way

.

Have a closer look

All terms contain 2 × 6 = 12 as factor

So, 2×6 ( 1+ 3 + 6 + 10 + 15..)

12( 1+ (1+2) +(1+2+3)+ (1+2+3+4)+.....)

12( (1+1+....n terms) + (2+2...(n-1) terms + (3+3+....(n-2) terms +......n )

12( n + 2(n-1) + 3(n-2) +.....n( n- ( n-1))

12( ( n+1) sigma(n) - sigma( n^2) ))

12[( n+1)[ n ( n+1)/2 - n( n+1)( 2n+1)/6)

= 12n ( n+1)[( n+1)/2 - ( 2n+1)/6)

= 12n( n+1) [ 3n +3 - 2n -2)/6

= 2 n( n+1)( n +1)

= 2n ( n+1)^2

#answerwithquality #BAL

Answer 2

Answer:

it is not a.p.

2*6+4*9+6*12+..........

common difference of first two term is 24.

common difference of second and third term is 36.

thus it is not right.

then we dont find it.

may be it help you.