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Find the nth term of this series  (1 \times 1)+(4 \times 3)+(9 \times 5)+(16 \times 7)+\ldots ..

Option: 1

n \times(2n-1)


Option: 2

(2n+1) \times(2n-1)


Option: 3

n^2 \times(2n-1)


Option: 4

2n \times(2n-1)


Answers (1)

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In this Question,

\\(\1 \times 1)+(4 \times 3)+(9 \times 5)+(16 \times 7)+\ldots ..\\ \\=(\1 \times 1)+(2^2 \times 3)+(3^2 \times 5)+(4^2 \times 7)+\ldots .. \\\\ \text{We can observe} \\\\ T_n=n^2 \times(2n-1)

 

Alternate method:

We can also check the options one by one if pattern is not clearly visible

The option that gives the correct first term by putting n=1, correct second term by putting n=2 and so on will be the correct option

Eg, for option (A) 

  • n=1 gives first term as 1x1
  • n=2 gives second term as 2x3, which is not same as the second term in the given sequence, hence option (A) is wrong

Similarly we can check other options as well

Posted by

Sanket Gandhi

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