Discussion Forum

1)

Let A be the sum of the first 20 terms and B be the sum of first 40 terms of the series

1²+2.2²+3²+2.4²+5²+2.6²+........  If B-2A=100λ  then λ is equal to

Answer:

Option B

Explanation:

We have

   1²+2.2²+3²+2.4²+5²+2.6²+  .......

A= Sum of first 20 terms

B= Sum of first 40terms

$\therefore$   A= 1²+2.2²+3²+2.4²+5²+2.6²+........+2.20²

A= (1²+2²+3²⁺.........+20²) + (2²+4²+6²+........+20²)

A= (1²+2²+3²+.......+20²)+4(1²+2²+3²+........+10²)

$A= \frac{20\times 21\times 41}{6}+\frac{4\times10 \times 11\times21}{6}$

$A= \frac{20\times 21}{6}(41+22)=\frac{20\times 41\times63}{6}$

Similarly , B= (1²+2²+3²+........+40²)+ 4 ( 1²+2²+3²+.......+20²)

      $B= \frac{40\times41\times81}{6}+\frac{4\times20\times21\times41}{6}$

     $B= \frac{40\times41}{6}(81+42)=\frac{40\times41\times123}{6}$

Now, B-2A=100λ

$\therefore$    $\frac{40\times41\times123}{6}-\frac{2\times20\times21\times63}{6}=100\lambda$

$\Rightarrow$        $\frac{40}{6}(5043-1323)=100\lambda$

$\Rightarrow$        $\frac{40}{6}\times 3720=100\lambda$

$\Rightarrow$        $40\times620=100\lambda$

$\Rightarrow$     $\lambda =\frac{40\times620}{100}=248$