Algenbra

Let X Remain UNKNOWN only

In this article, we explain the non-conventional way to solve Algebra questions without formulating equations. These alternative methods will enable you to save a lot of time as compared to the traditional and conventional approaches. Further, using these methods you will avoid making silly errors that creep in when the calculation work is lengthy.

You will understand how to solve Algebra questions without ‘x’ with the help of following examples:

Example 1:

The sum to n terms of the infinite series 1. 3² + 2. 5² + 3. 7² +....is:

1.  n/6 (n+1)(6n² + 14n + 7)
2.  n/6 (n+1)(2n + 1)(3n + 1)
3.  4n³ + 4n² + n
4. None of these

Solution:

The series in the question is obtained by multiplying 1, 2, 3, 4, … and 3², 5², 7², 9², … i.e. all natural number starting with 1, and the squares of all odd natural numbers, except one. Here, you need to find the general formula that will hold true for all the values of n.

Now, for the formula to satisfy all values of n, it should be applicable for n = 1 and n = 2 also. You can see that the first term in the given series is 9, and the second term is 50. This means that the sum of first two terms will be 9 + 50 = 59. So, that option in which you get 9, for n = 1 and 59, for n = 2 will be the answer. Here, the 1^st option gives the required values, and hence is the answer.

Note: n = 1 will be used for rejection or elimination of options only (as two different series can start with same number), and n = 2 will confirm the answer.

Example 2:

The sum to n terms of the series 1/2 + 3/4 + 7/8 + 15/16 +........ is equal to

1. 2ⁿ - n - 1                    2.  1 - 2^-n
3. 2^-n + n - 1                  4.  2ⁿ - 1

Solution:

Here, the first term is 1/2 and sum of the first two terms is 1/2  + 3/4  = 5/4 . So, by taking n = 1, you should get ½ as the answer, and by taking n = 2, you should get 5/4 as the answer. Here, the third option satisfies both of these and hence it is the answer.

However, you should be careful as n = 1 may satisfy one more of the options. This condition should be used only for rejection or elimination of options and the answer should always be confirmed by taking n = 2.

Example 3:

The nth term of the series 1+ 4/5 + 7/5² + 10/5³+.... will be:

1. 3n+1/5^n-1                  2. 3n - 2/5^n-1
3. 3n - 1/5ⁿ                    4. 3n+2/5^n-1

Solution:

By taking n = 1, we should get the 1^st term i.e. 1, and by taking n = 2, we should get the second term i.e. 4/5. Here, only the second option satisfies both of those and hence it is the answer.