2. 13+23+33+ … +n3= <math> <mrow> <msup> <mrow> <mrow> <mo>(</mo> <mrow> <mfrac> <mrow> <mi>n</mi> <mo stretchy="false">(</mo> <mi>n</mi> <mo>+</mo> <mn>1</mn> <mo stretchy="false">)</mo> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </math>

2. Let the given statement be P(n) i.e.,P(n)=13+23+33+ … +n3= (n(n+1)2)2For, n=1, P(n)=13=1= (1(1+1)2)2=(1.22)2=12=1which is true.Consider P(k) be true for some positive integer k13+23+33+ … +k3= (k(k+1)2)2 ---------- (1)Now, let us prove that P(k+1) is true.Here, 13+23+33+ … +k3+(k+1)3By using eq (1)= (k(k+1)2)2+(k+1)3= k2(k+1)2+4·(k+1)34= (k+1)2[k2+4(k+1)]4= (k+1)2{k2+4k+4}4=(k+1)2(k+2)24= {(k+1)(k+1+1)2}2? P(k+1) is true whenever P(k) is true.Hence, from the principle of mathematical induction, P(n) is true for all natural numbers n.

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Ncert Solutions Maths class 11th Maths Class 11th Principle of Mathematical Induction

2. 1³+2³+3³+ … +n³= ${(\frac{n (n + 1)}{2})}^{2}$

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Payal Gupta | Contributor-Level 10

4 months ago

2. Let the given statement be P(n) i.e.,

P(n)=1³+2³+3³+ … +n³= ${(\frac{n (n + 1)}{2})}^{2}$

For, n=1, P(n)=1³=1= ${(\frac{1 (1 + 1)}{2})}^{2} = {(\frac{1.2}{2})}^{2} = 1^{2} = 1$

which is true.

Consider P(k) be true for some positive integer k

1³+2³+3³+ … +k³= ${(\frac{k (k + 1)}{2})}^{2}$ ---------- (1)

Now, let us prove that P(k+1) is true.

Here, 1³+2³+3³+ … +k³+(k+1)³

By using eq (1)

= ${(\frac{k (k + 1)}{2})}^{2} + {(k + 1)}^{3}$

= $\frac{k^{2} (k + 1)^{2} + 4 \cdot {(k + 1)}^{3}}{4}$

= $\frac{{(k + 1)}^{2} [k^{2} + 4 (k + 1)]}{4}$

= $\frac{{(k + 1)}^{2} {k^{2} + 4 k + 4}}{4} = \frac{{(k + 1)}^{2} (k + 2)^{2}}{4}$

= ${\frac{(k + 1) (k + 1 + 1)}{2}}^{2}$

? P(k+1) is true whenever P(k) is true.

Hence, from the principle of mathematical induction, P(n) is true for all natural numbers n.

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24. Let P (n) be the statement “ 2n+7< (n+3)2”

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$∴$ By the principle of mathematical induction, P (n)is true for all n $\in$ N.

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