5 Marks Questions

Question

Match the questions given under Column I with their appropriate answers given under the Column II. Column I Column II (a) 1^2+2^2+3^2+....+n^2 (i) [ n(n+1) 2 ]^2 (b) 1^3+2^3+3^3+....n^3 (ii) n(n+1) (c) 2+4+6+....+2n (iii) n(n+1)(2n+1) 6 (d) 1+2+3+....n (iv) n(n+1) 2

Vidyadip · Accepted Answer

Column I Column II (a) 1^2+2^2+3^2+....+n^2 (iii) n(n+1)(2n+1) 6 (b) 1^3+2^3+3^3+....n^3 (i) [ n(n+1) 2 ]^2 (c) 2+4+6+....+2n (ii) n(n+1) (d) 1+2+3+....n (iv) n(n+1) 2 Solution: 1. Let S=1^2+2^2+3^2+....+n^2 We have, n^3-(n-1)^3=3n^2-3n+1; And by changing n into n - 1, (n-1)^3-(n-2)^3=3(n-1)^2-3(n-1)+1; (n-2)^2-(n-3)^3=3(n-2)^2-3(n-2)+1; .......... .......... .......... 3^3-2^3=3.3^2-3.3+1; 2^3-1^2=3.2^2-3.2+1; 1^2-0^2=3.1^2-3.1+1 Hence, by addition, n^3=3(1^2+2^2+3^2+....+n^2)-3(1+2+3+....+n)+n =3s- 3n(n+1) 2 +n 3S=n^3-n+ 3n(n+1) 2 =n(n+1) (n-1+3 2 ) = n(n+1)(2n+1) 6 2. Let S=1^3+2^3+3^3+.....+n^3 We have, n^4-(n-1)^4=4n^3-6n^2+4n-1; (n-1)^4-(n-2)^4=4(n-1)^3-6(n-1)^2+4(n-1)-1; (n-2)^4-(n-3)^4=4(n-2)^3-6(n-2)^2+4(n-2)-1; .......... .......... .......... 3^4-2^4=4.3^3-6.3^2+4.3-1; 2^4-1^4=4.2^3-6.2^2+4.2-1; 1^4-0^4=4.1^3-6.1^2+4.1-1. Hence, by addition, n^4=4S-6 (1^2+2^2+....+n^2 )+4(1+2+....+n)-n; 4S=n^4+n+6 (1^2+2^2+....+n^2 )-4(1+2+....+n) =n^4+n+n(n+1)(2n+1)-2n(n+1) =n( n+1 ) (n^2-n+1+2n+1-2 ) =n(n+1) (n^2+n ) S= n^2(n+1)^2 4 = n(n+1) 2 ^2 3. 1+2+3+....+n= n(n+2) 2 2+4+6+....+2n=2(1+2+3+....+n) =2 n 2 (1+n)=n(n+1) 4. 1 + 2 + 3 + .... + n = Sum of n terms of A.P. with first term '1' and common difference '1'= n 2 (1+n)

	Column I		Column II
$(a)$	$1^2+2^2+3^2+....+\text{n}^2$	$(i)$	$\Big[\frac{\text{n}(\text{n}+1)}{2}\Big]^2$
$(b)$	$1^3+2^3+3^3+....\text{n}^3$	$(ii)$	$\text{n}(\text{n}+1)$
$(c)$	$2+4+6+....+2\text{n}$	$(iii)$	$\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{6}$
$(d)$	$1+2+3+....\text{n}$	$(iv)$	$\frac{\text{n}(\text{n}+1)}{2}$

	Column I		Column II
$(a)$	$1^2+2^2+3^2+....+\text{n}^2$	$(iii)$	$\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{6}$
$(b)$	$1^3+2^3+3^3+....\text{n}^3$	$(i)$	$\Big[\frac{\text{n}(\text{n}+1)}{2}\Big]^2$
$(c)$	$2+4+6+....+2\text{n}$	$(ii)$	$\text{n}(\text{n}+1)$
$(d)$	$1+2+3+....\text{n}$	$(iv)$	$\frac{\text{n}(\text{n}+1)}{2}$

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