Question

prove that any odd square (2n+1)^2 is equal to sum of n terms of an AP increased by unity.​

Answer 1

solution:-

The difference between any two consecutive square numbers,

where n is the square root of the smaller one, is

(n+1)^2 - n^2

= (n^2 + 2n + 1) - n^2

= 2n + 1

so the sum of all the odd numbers from 1 to 2n+1, for any positive integer n,

is the square of n+1.

If we add "unity" (that is, the number 1) to the sum of any number of terms of the arithmetic progression

3, 5, 7, 9, ...

we get

1 + 3 + 5 + 7 + 9 + ... + (2n+1)

where n is the number of terms from the A.P. (which started with 3).

And we have seen that this sum is equal to the square of n+1.

Answer 2

Step-by-step explanation:

We want to prove that ∑i=1n(2i−1)=n2∑i=1n(2i−1)=n2. For this, we could do a proof by induction. The general premise of a proof by induction is that we prove that it is true for one value, called the base case or basis (typically n=1n=1), then we prove that it is is true for any number plus one. Thus, since we have proved that it is true for n=1n=1, then we know that it is true for n=2n=2, so we know that it is true for n=3n=3, and so on.

Prove that the formula is true for n=1n=1.

∑i=11(2i−1)=12∑i=11(2i−1)=12

⟹2(1)−1=1⟹2(1)−1=1

⟹1=1✓⟹1=1✓

Assume that the formula is true for n=kn=k.

∑i=1k(2i−1)=k2∑i=1k(2i−1)=k2

Prove that the formula is true for n=k+1n=k+1.

∑i=1k+1(2i−1)=(k+1)2∑i=1k+1(2i−1)=(k+1)2

By the definition of summation,

⟹∑i=1k(2i−1)+(2(k+1)−1)=(k+1)2⟹∑i=1k(2i−1)+(2(k+1)−1)=(k+1)2

Because of our assumption in step 2, we get

⟹k2+(2(k+1)−1)=(k+1)2⟹k2+(2(k+1)−1)=(k+1)2

⟹k2+2k+2−1=(k+1)2⟹k2+2k+2−1=(k+1)2

⟹k2+2k+1=(k+1)2⟹k2+2k+1=(k+1)2

⟹(k+1)2=(k+1)2✓⟹(k+1)2=(k+1)2✓

Thus, we have proved that ∑i=1n(2i−1)=n2∑i=1n(2i−1)=n2.