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2026-01-01
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2026-02-28
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<p>Here, in this section, we will learn how to derive the formula to find the sum of squares like 12 + 22 + 32 + … </p>
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<p>Here, in this section, we will learn how to derive the formula to find the sum of squares like 12 + 22 + 32 + … </p>
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<p>The formula used is 12 + 22 + 32 + … + n2 = n (n+1) (2n +1) / 6 </p>
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<p>The formula used is 12 + 22 + 32 + … + n2 = n (n+1) (2n +1) / 6 </p>
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<p>To see how the formula is derived, we use of the process called<a>mathematical induction</a>. It makes sure that the formula works for one number, proving that it applies to other numbers as well. Follow the steps given below:-</p>
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<p>To see how the formula is derived, we use of the process called<a>mathematical induction</a>. It makes sure that the formula works for one number, proving that it applies to other numbers as well. Follow the steps given below:-</p>
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<p><strong>Step 1:</strong>Check if the formula works for number 1. Therefore, we can take the value of ‘n’ as 1. </p>
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<p><strong>Step 1:</strong>Check if the formula works for number 1. Therefore, we can take the value of ‘n’ as 1. </p>
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<p>LHS is 12 = 1</p>
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<p>LHS is 12 = 1</p>
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<p>RHS = n (n+1) (2n +1) / 6 </p>
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<p>RHS = n (n+1) (2n +1) / 6 </p>
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<p>Now substituting the value of ‘n’ as 1 in the formula, we get 1 (1+1) ((2 x 1) +1) / 6 = 1 x 2 x 36 = 6/6 = 1</p>
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<p>Now substituting the value of ‘n’ as 1 in the formula, we get 1 (1+1) ((2 x 1) +1) / 6 = 1 x 2 x 36 = 6/6 = 1</p>
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<p>Since both the LHS and RHS are the same, we can say that the formula works for n = 1</p>
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<p>Since both the LHS and RHS are the same, we can say that the formula works for n = 1</p>
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<p><strong>Step 2:</strong>Now let’s check if the formula works for the number ‘k’. Assume that,</p>
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<p><strong>Step 2:</strong>Now let’s check if the formula works for the number ‘k’. Assume that,</p>
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<p>12 + 22 + 33 + … + k2 = k(k+1) (2k + 1) / 6</p>
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<p>12 + 22 + 33 + … + k2 = k(k+1) (2k + 1) / 6</p>
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<p>Since we are assuming for number ‘k’ to be true, this process is called the inductive hypothesis. Now we have to check if it works for the number ‘k + 1’</p>
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<p>Since we are assuming for number ‘k’ to be true, this process is called the inductive hypothesis. Now we have to check if it works for the number ‘k + 1’</p>
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<p><strong>Step 3</strong>: Let’s prove that the formula works for ‘k + 1’ 12 + 22 + 33 + … + (k + 1)2 = (k+1) (k+2) (2k+3) / 6</p>
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<p><strong>Step 3</strong>: Let’s prove that the formula works for ‘k + 1’ 12 + 22 + 33 + … + (k + 1)2 = (k+1) (k+2) (2k+3) / 6</p>
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<p>The sum of squares up to k is 12 + 22 + 33 + … + k2 = k(k+1) (2k + 1) / 6</p>
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<p>The sum of squares up to k is 12 + 22 + 33 + … + k2 = k(k+1) (2k + 1) / 6</p>
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<p>Now we can add (k + 1)2 to both sides:-</p>
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<p>Now we can add (k + 1)2 to both sides:-</p>
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<p>12 + 22 + 32 + … + k2 + (k + 1)2 = k(k+1) (2k + 1) / 6 + (k + 1)2</p>
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<p>12 + 22 + 32 + … + k2 + (k + 1)2 = k(k+1) (2k + 1) / 6 + (k + 1)2</p>
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<p><strong>Step 4:</strong>Take out the<a>common factor</a>(k + 1) from both side k+1/ 6 [k (2k + 1) + 6 (k +1)] By simplifying the brackets, we get:-</p>
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<p><strong>Step 4:</strong>Take out the<a>common factor</a>(k + 1) from both side k+1/ 6 [k (2k + 1) + 6 (k +1)] By simplifying the brackets, we get:-</p>
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<p>k+1/ 6 (2k2 + k + 6k + 6) = k+1/ 6 (2k2 + 7k + 6)</p>
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<p>k+1/ 6 (2k2 + k + 6k + 6) = k+1/ 6 (2k2 + 7k + 6)</p>
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<p><strong>Step 5:</strong>Now factor (k+2) from (2k2 + 7k + 6) by splitting them By doing so, we get:- k+1 6 [2k (k + 2) + 3 (k + 2)] Factoring out (k + 2), we get:-</p>
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<p><strong>Step 5:</strong>Now factor (k+2) from (2k2 + 7k + 6) by splitting them By doing so, we get:- k+1 6 [2k (k + 2) + 3 (k + 2)] Factoring out (k + 2), we get:-</p>
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<p>k+1 6 (k + 2) (2k + 3)</p>
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<p>k+1 6 (k + 2) (2k + 3)</p>
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<p><strong>Step 6:</strong>We get the final answer as:- (k+1) (k + 2) (2k + 3) / 6</p>
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<p><strong>Step 6:</strong>We get the final answer as:- (k+1) (k + 2) (2k + 3) / 6</p>
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<p>Since the formula is true for n=1, and we've shown that if it works for any number k, it also works for k+1, we can conclude that the formula is true for all natural numbers. Therefore, the sum of squares formula is:</p>
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<p>Since the formula is true for n=1, and we've shown that if it works for any number k, it also works for k+1, we can conclude that the formula is true for all natural numbers. Therefore, the sum of squares formula is:</p>
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<p>S = 12 + 22 + 32 + 42 + …. + n2 = n (n+1) (2n +1) / 6 </p>
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<p>S = 12 + 22 + 32 + 42 + …. + n2 = n (n+1) (2n +1) / 6 </p>
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